/*
 * Bignum routines for RSA and DH and stuff.
 */

#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

#include "misc.h"

#include "sshbn.h"

#define BIGNUM_INTERNAL
typedef BignumInt *Bignum;

#include "ssh.h"

BignumInt bnZero[1] = { 0 };
BignumInt bnOne[2] = { 1, 1 };

/*
 * The Bignum format is an array of `BignumInt'. The first
 * element of the array counts the remaining elements. The
 * remaining elements express the actual number, base 2^BIGNUM_INT_BITS, _least_
 * significant digit first. (So it's trivial to extract the bit
 * with value 2^n for any n.)
 *
 * All Bignums in this module are positive. Negative numbers must
 * be dealt with outside it.
 *
 * INVARIANT: the most significant word of any Bignum must be
 * nonzero.
 */

Bignum Zero = bnZero, One = bnOne;

static Bignum newbn(int length)
{
    Bignum b;

    assert(length >= 0 && length < INT_MAX / BIGNUM_INT_BITS);

    b = snewn(length + 1, BignumInt);
    if (!b)
	abort();		       /* FIXME */
    memset(b, 0, (length + 1) * sizeof(*b));
    b[0] = length;
    return b;
}

void bn_restore_invariant(Bignum b)
{
    while (b[0] > 1 && b[b[0]] == 0)
	b[0]--;
}

Bignum copybn(Bignum orig)
{
    Bignum b = snewn(orig[0] + 1, BignumInt);
    if (!b)
	abort();		       /* FIXME */
    memcpy(b, orig, (orig[0] + 1) * sizeof(*b));
    return b;
}

void freebn(Bignum b)
{
    /*
     * Burn the evidence, just in case.
     */
    smemclr(b, sizeof(b[0]) * (b[0] + 1));
    sfree(b);
}

Bignum bn_power_2(int n)
{
    Bignum ret;

    assert(n >= 0);

    ret = newbn(n / BIGNUM_INT_BITS + 1);
    bignum_set_bit(ret, n, 1);
    return ret;
}

/*
 * Internal addition. Sets c = a - b, where 'a', 'b' and 'c' are all
 * big-endian arrays of 'len' BignumInts. Returns a BignumInt carried
 * off the top.
 */
static BignumInt internal_add(const BignumInt *a, const BignumInt *b,
                              BignumInt *c, int len)
{
    int i;
    BignumDblInt carry = 0;

    for (i = len-1; i >= 0; i--) {
        carry += (BignumDblInt)a[i] + b[i];
        c[i] = (BignumInt)carry;
        carry >>= BIGNUM_INT_BITS;
    }

    return (BignumInt)carry;
}

/*
 * Internal subtraction. Sets c = a - b, where 'a', 'b' and 'c' are
 * all big-endian arrays of 'len' BignumInts. Any borrow from the top
 * is ignored.
 */
static void internal_sub(const BignumInt *a, const BignumInt *b,
                         BignumInt *c, int len)
{
    int i;
    BignumDblInt carry = 1;

    for (i = len-1; i >= 0; i--) {
        carry += (BignumDblInt)a[i] + (b[i] ^ BIGNUM_INT_MASK);
        c[i] = (BignumInt)carry;
        carry >>= BIGNUM_INT_BITS;
    }
}

/*
 * Compute c = a * b.
 * Input is in the first len words of a and b.
 * Result is returned in the first 2*len words of c.
 *
 * 'scratch' must point to an array of BignumInt of size at least
 * mul_compute_scratch(len). (This covers the needs of internal_mul
 * and all its recursive calls to itself.)
 */
#define KARATSUBA_THRESHOLD 50
static int mul_compute_scratch(int len)
{
    int ret = 0;
    while (len > KARATSUBA_THRESHOLD) {
        int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
        int midlen = botlen + 1;
        ret += 4*midlen;
        len = midlen;
    }
    return ret;
}
static void internal_mul(const BignumInt *a, const BignumInt *b,
			 BignumInt *c, int len, BignumInt *scratch)
{
    if (len > KARATSUBA_THRESHOLD) {
        int i;

        /*
         * Karatsuba divide-and-conquer algorithm. Cut each input in
         * half, so that it's expressed as two big 'digits' in a giant
         * base D:
         *
         *   a = a_1 D + a_0
         *   b = b_1 D + b_0
         *
         * Then the product is of course
         *
         *  ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
         *
         * and we compute the three coefficients by recursively
         * calling ourself to do half-length multiplications.
         *
         * The clever bit that makes this worth doing is that we only
         * need _one_ half-length multiplication for the central
         * coefficient rather than the two that it obviouly looks
         * like, because we can use a single multiplication to compute
         *
         *   (a_1 + a_0) (b_1 + b_0) = a_1 b_1 + a_1 b_0 + a_0 b_1 + a_0 b_0
         *
         * and then we subtract the other two coefficients (a_1 b_1
         * and a_0 b_0) which we were computing anyway.
         *
         * Hence we get to multiply two numbers of length N in about
         * three times as much work as it takes to multiply numbers of
         * length N/2, which is obviously better than the four times
         * as much work it would take if we just did a long
         * conventional multiply.
         */

        int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
        int midlen = botlen + 1;
        BignumDblInt carry;
#ifdef KARA_DEBUG
        int i;
#endif

        /*
         * The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping
         * in the output array, so we can compute them immediately in
         * place.
         */

#ifdef KARA_DEBUG
        printf("a1,a0 = 0x");
        for (i = 0; i < len; i++) {
            if (i == toplen) printf(", 0x");
            printf("%0*x", BIGNUM_INT_BITS/4, a[i]);
        }
        printf("\n");
        printf("b1,b0 = 0x");
        for (i = 0; i < len; i++) {
            if (i == toplen) printf(", 0x");
            printf("%0*x", BIGNUM_INT_BITS/4, b[i]);
        }
        printf("\n");
#endif

        /* a_1 b_1 */
        internal_mul(a, b, c, toplen, scratch);
#ifdef KARA_DEBUG
        printf("a1b1 = 0x");
        for (i = 0; i < 2*toplen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
        }
        printf("\n");
#endif

        /* a_0 b_0 */
        internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen, scratch);
#ifdef KARA_DEBUG
        printf("a0b0 = 0x");
        for (i = 0; i < 2*botlen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, c[2*toplen+i]);
        }
        printf("\n");
#endif

        /* Zero padding. midlen exceeds toplen by at most 2, so just
         * zero the first two words of each input and the rest will be
         * copied over. */
        scratch[0] = scratch[1] = scratch[midlen] = scratch[midlen+1] = 0;

        for (i = 0; i < toplen; i++) {
            scratch[midlen - toplen + i] = a[i]; /* a_1 */
            scratch[2*midlen - toplen + i] = b[i]; /* b_1 */
        }

        /* compute a_1 + a_0 */
        scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);
#ifdef KARA_DEBUG
        printf("a1plusa0 = 0x");
        for (i = 0; i < midlen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
        }
        printf("\n");
#endif
        /* compute b_1 + b_0 */
        scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,
                                       scratch+midlen+1, botlen);
#ifdef KARA_DEBUG
        printf("b1plusb0 = 0x");
        for (i = 0; i < midlen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, scratch[midlen+i]);
        }
        printf("\n");
#endif

        /*
         * Now we can do the third multiplication.
         */
        internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen,
                     scratch + 4*midlen);
#ifdef KARA_DEBUG
        printf("a1plusa0timesb1plusb0 = 0x");
        for (i = 0; i < 2*midlen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
        }
        printf("\n");
#endif

        /*
         * Now we can reuse the first half of 'scratch' to compute the
         * sum of the outer two coefficients, to subtract from that
         * product to obtain the middle one.
         */
        scratch[0] = scratch[1] = scratch[2] = scratch[3] = 0;
        for (i = 0; i < 2*toplen; i++)
            scratch[2*midlen - 2*toplen + i] = c[i];
        scratch[1] = internal_add(scratch+2, c + 2*toplen,
                                  scratch+2, 2*botlen);
#ifdef KARA_DEBUG
        printf("a1b1plusa0b0 = 0x");
        for (i = 0; i < 2*midlen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
        }
        printf("\n");
#endif

        internal_sub(scratch + 2*midlen, scratch,
                     scratch + 2*midlen, 2*midlen);
#ifdef KARA_DEBUG
        printf("a1b0plusa0b1 = 0x");
        for (i = 0; i < 2*midlen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
        }
        printf("\n");
#endif

        /*
         * And now all we need to do is to add that middle coefficient
         * back into the output. We may have to propagate a carry
         * further up the output, but we can be sure it won't
         * propagate right the way off the top.
         */
        carry = internal_add(c + 2*len - botlen - 2*midlen,
                             scratch + 2*midlen,
                             c + 2*len - botlen - 2*midlen, 2*midlen);
        i = 2*len - botlen - 2*midlen - 1;
        while (carry) {
            assert(i >= 0);
            carry += c[i];
            c[i] = (BignumInt)carry;
            carry >>= BIGNUM_INT_BITS;
            i--;
        }
#ifdef KARA_DEBUG
        printf("ab = 0x");
        for (i = 0; i < 2*len; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
        }
        printf("\n");
#endif

    } else {
        int i;
        BignumInt carry;
        BignumDblInt t;
        const BignumInt *ap, *bp;
        BignumInt *cp, *cps;

        /*
         * Multiply in the ordinary O(N^2) way.
         */

        for (i = 0; i < 2 * len; i++)
            c[i] = 0;

        for (cps = c + 2*len, ap = a + len; ap-- > a; cps--) {
            carry = 0;
            for (cp = cps, bp = b + len; cp--, bp-- > b ;) {
                t = (MUL_WORD(*ap, *bp) + carry) + *cp;
                *cp = (BignumInt) t;
                carry = (BignumInt)(t >> BIGNUM_INT_BITS);
            }
            *cp = carry;
        }
    }
}

/*
 * Variant form of internal_mul used for the initial step of
 * Montgomery reduction. Only bothers outputting 'len' words
 * (everything above that is thrown away).
 */
static void internal_mul_low(const BignumInt *a, const BignumInt *b,
                             BignumInt *c, int len, BignumInt *scratch)
{
    if (len > KARATSUBA_THRESHOLD) {
        int i;

        /*
         * Karatsuba-aware version of internal_mul_low. As before, we
         * express each input value as a shifted combination of two
         * halves:
         *
         *   a = a_1 D + a_0
         *   b = b_1 D + b_0
         *
         * Then the full product is, as before,
         *
         *  ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
         *
         * Provided we choose D on the large side (so that a_0 and b_0
         * are _at least_ as long as a_1 and b_1), we don't need the
         * topmost term at all, and we only need half of the middle
         * term. So there's no point in doing the proper Karatsuba
         * optimisation which computes the middle term using the top
         * one, because we'd take as long computing the top one as
         * just computing the middle one directly.
         *
         * So instead, we do a much more obvious thing: we call the
         * fully optimised internal_mul to compute a_0 b_0, and we
         * recursively call ourself to compute the _bottom halves_ of
         * a_1 b_0 and a_0 b_1, each of which we add into the result
         * in the obvious way.
         *
         * In other words, there's no actual Karatsuba _optimisation_
         * in this function; the only benefit in doing it this way is
         * that we call internal_mul proper for a large part of the
         * work, and _that_ can optimise its operation.
         */

        int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */

        /*
         * Scratch space for the various bits and pieces we're going
         * to be adding together: we need botlen*2 words for a_0 b_0
         * (though we may end up throwing away its topmost word), and
         * toplen words for each of a_1 b_0 and a_0 b_1. That adds up
         * to exactly 2*len.
         */

        /* a_0 b_0 */
        internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen,
                     scratch + 2*len);

        /* a_1 b_0 */
        internal_mul_low(a, b + len - toplen, scratch + toplen, toplen,
                         scratch + 2*len);

        /* a_0 b_1 */
        internal_mul_low(a + len - toplen, b, scratch, toplen,
                         scratch + 2*len);

        /* Copy the bottom half of the big coefficient into place */
        for (i = 0; i < botlen; i++)
            c[toplen + i] = scratch[2*toplen + botlen + i];

        /* Add the two small coefficients, throwing away the returned carry */
        internal_add(scratch, scratch + toplen, scratch, toplen);

        /* And add that to the large coefficient, leaving the result in c. */
        internal_add(scratch, scratch + 2*toplen + botlen - toplen,
                     c, toplen);

    } else {
        int i;
        BignumInt carry;
        BignumDblInt t;
        const BignumInt *ap, *bp;
        BignumInt *cp, *cps;

        /*
         * Multiply in the ordinary O(N^2) way.
         */

        for (i = 0; i < len; i++)
            c[i] = 0;

        for (cps = c + len, ap = a + len; ap-- > a; cps--) {
            carry = 0;
            for (cp = cps, bp = b + len; bp--, cp-- > c ;) {
                t = (MUL_WORD(*ap, *bp) + carry) + *cp;
                *cp = (BignumInt) t;
                carry = (BignumInt)(t >> BIGNUM_INT_BITS);
            }
        }
    }
}

/*
 * Montgomery reduction. Expects x to be a big-endian array of 2*len
 * BignumInts whose value satisfies 0 <= x < rn (where r = 2^(len *
 * BIGNUM_INT_BITS) is the Montgomery base). Returns in the same array
 * a value x' which is congruent to xr^{-1} mod n, and satisfies 0 <=
 * x' < n.
 *
 * 'n' and 'mninv' should be big-endian arrays of 'len' BignumInts
 * each, containing respectively n and the multiplicative inverse of
 * -n mod r.
 *
 * 'tmp' is an array of BignumInt used as scratch space, of length at
 * least 3*len + mul_compute_scratch(len).
 */
static void monty_reduce(BignumInt *x, const BignumInt *n,
                         const BignumInt *mninv, BignumInt *tmp, int len)
{
    int i;
    BignumInt carry;

    /*
     * Multiply x by (-n)^{-1} mod r. This gives us a value m such
     * that mn is congruent to -x mod r. Hence, mn+x is an exact
     * multiple of r, and is also (obviously) congruent to x mod n.
     */
    internal_mul_low(x + len, mninv, tmp, len, tmp + 3*len);

    /*
     * Compute t = (mn+x)/r in ordinary, non-modular, integer
     * arithmetic. By construction this is exact, and is congruent mod
     * n to x * r^{-1}, i.e. the answer we want.
     *
     * The following multiply leaves that answer in the _most_
     * significant half of the 'x' array, so then we must shift it
     * down.
     */
    internal_mul(tmp, n, tmp+len, len, tmp + 3*len);
    carry = internal_add(x, tmp+len, x, 2*len);
    for (i = 0; i < len; i++)
        x[len + i] = x[i], x[i] = 0;

    /*
     * Reduce t mod n. This doesn't require a full-on division by n,
     * but merely a test and single optional subtraction, since we can
     * show that 0 <= t < 2n.
     *
     * Proof:
     *  + we computed m mod r, so 0 <= m < r.
     *  + so 0 <= mn < rn, obviously
     *  + hence we only need 0 <= x < rn to guarantee that 0 <= mn+x < 2rn
     *  + yielding 0 <= (mn+x)/r < 2n as required.
     */
    if (!carry) {
        for (i = 0; i < len; i++)
            if (x[len + i] != n[i])
                break;
    }
    if (carry || i >= len || x[len + i] > n[i])
        internal_sub(x+len, n, x+len, len);
}

static void internal_add_shifted(BignumInt *number,
				 BignumInt n, int shift)
{
    int word = 1 + (shift / BIGNUM_INT_BITS);
    int bshift = shift % BIGNUM_INT_BITS;
    BignumDblInt addend;

    addend = (BignumDblInt)n << bshift;

    while (addend) {
        assert(word <= number[0]);
	addend += number[word];
	number[word] = (BignumInt) addend & BIGNUM_INT_MASK;
	addend >>= BIGNUM_INT_BITS;
	word++;
    }
}

/*
 * Compute a = a % m.
 * Input in first alen words of a and first mlen words of m.
 * Output in first alen words of a
 * (of which first alen-mlen words will be zero).
 * The MSW of m MUST have its high bit set.
 * Quotient is accumulated in the `quotient' array, which is a Bignum
 * rather than the internal bigendian format. Quotient parts are shifted
 * left by `qshift' before adding into quot.
 */
static void internal_mod(BignumInt *a, int alen,
			 BignumInt *m, int mlen,
			 BignumInt *quot, int qshift)
{
    BignumInt m0, m1, h;
    int i, k;

    m0 = m[0];
    assert(m0 >> (BIGNUM_INT_BITS-1) == 1);
    if (mlen > 1)
	m1 = m[1];
    else
	m1 = 0;

    for (i = 0; i <= alen - mlen; i++) {
	BignumDblInt t;
        BignumInt q, r, c, ai1;

	if (i == 0) {
	    h = 0;
	} else {
	    h = a[i - 1];
	    a[i - 1] = 0;
	}

	if (i == alen - 1)
	    ai1 = 0;
	else
	    ai1 = a[i + 1];

	/* Find q = h:a[i] / m0 */
	if (h >= m0) {
	    /*
	     * Special case.
	     * 
	     * To illustrate it, suppose a BignumInt is 8 bits, and
	     * we are dividing (say) A1:23:45:67 by A1:B2:C3. Then
	     * our initial division will be 0xA123 / 0xA1, which
	     * will give a quotient of 0x100 and a divide overflow.
	     * However, the invariants in this division algorithm
	     * are not violated, since the full number A1:23:... is
	     * _less_ than the quotient prefix A1:B2:... and so the
	     * following correction loop would have sorted it out.
	     * 
	     * In this situation we set q to be the largest
	     * quotient we _can_ stomach (0xFF, of course).
	     */
	    q = BIGNUM_INT_MASK;
	} else {
	    /* Macro doesn't want an array subscript expression passed
	     * into it (see definition), so use a temporary. */
	    BignumInt tmplo = a[i];
	    DIVMOD_WORD(q, r, h, tmplo, m0);

	    /* Refine our estimate of q by looking at
	     h:a[i]:a[i+1] / m0:m1 */
	    t = MUL_WORD(m1, q);
	    if (t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) {
		q--;
		t -= m1;
		r = (r + m0) & BIGNUM_INT_MASK;     /* overflow? */
		if (r >= (BignumDblInt) m0 &&
		    t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) q--;
	    }
	}

	/* Subtract q * m from a[i...] */
	c = 0;
	for (k = mlen - 1; k >= 0; k--) {
	    t = MUL_WORD(q, m[k]);
	    t += c;
	    c = (BignumInt)(t >> BIGNUM_INT_BITS);
	    if ((BignumInt) t > a[i + k])
		c++;
	    a[i + k] -= (BignumInt) t;
	}

	/* Add back m in case of borrow */
	if (c != h) {
	    t = 0;
	    for (k = mlen - 1; k >= 0; k--) {
		t += m[k];
		t += a[i + k];
		a[i + k] = (BignumInt) t;
		t = t >> BIGNUM_INT_BITS;
	    }
	    q--;
	}
	if (quot)
	    internal_add_shifted(quot, q, qshift + BIGNUM_INT_BITS * (alen - mlen - i));
    }
}

/*
 * Compute (base ^ exp) % mod, the pedestrian way.
 */
Bignum modpow_simple(Bignum base_in, Bignum exp, Bignum mod)
{
    BignumInt *a, *b, *n, *m, *scratch;
    int mshift;
    int mlen, scratchlen, i, j;
    Bignum base, result;

    /*
     * The most significant word of mod needs to be non-zero. It
     * should already be, but let's make sure.
     */
    assert(mod[mod[0]] != 0);

    /*
     * Make sure the base is smaller than the modulus, by reducing
     * it modulo the modulus if not.
     */
    base = bigmod(base_in, mod);

    /* Allocate m of size mlen, copy mod to m */
    /* We use big endian internally */
    mlen = mod[0];
    m = snewn(mlen, BignumInt);
    for (j = 0; j < mlen; j++)
	m[j] = mod[mod[0] - j];

    /* Shift m left to make msb bit set */
    for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
	if ((m[0] << mshift) & BIGNUM_TOP_BIT)
	    break;
    if (mshift) {
	for (i = 0; i < mlen - 1; i++)
	    m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
	m[mlen - 1] = m[mlen - 1] << mshift;
    }

    /* Allocate n of size mlen, copy base to n */
    n = snewn(mlen, BignumInt);
    i = mlen - base[0];
    for (j = 0; j < i; j++)
	n[j] = 0;
    for (j = 0; j < (int)base[0]; j++)
	n[i + j] = base[base[0] - j];

    /* Allocate a and b of size 2*mlen. Set a = 1 */
    a = snewn(2 * mlen, BignumInt);
    b = snewn(2 * mlen, BignumInt);
    for (i = 0; i < 2 * mlen; i++)
	a[i] = 0;
    a[2 * mlen - 1] = 1;

    /* Scratch space for multiplies */
    scratchlen = mul_compute_scratch(mlen);
    scratch = snewn(scratchlen, BignumInt);

    /* Skip leading zero bits of exp. */
    i = 0;
    j = BIGNUM_INT_BITS-1;
    while (i < (int)exp[0] && (exp[exp[0] - i] & ((BignumInt)1 << j)) == 0) {
	j--;
	if (j < 0) {
	    i++;
	    j = BIGNUM_INT_BITS-1;
	}
    }

    /* Main computation */
    while (i < (int)exp[0]) {
	while (j >= 0) {
	    internal_mul(a + mlen, a + mlen, b, mlen, scratch);
	    internal_mod(b, mlen * 2, m, mlen, NULL, 0);
	    if ((exp[exp[0] - i] & ((BignumInt)1 << j)) != 0) {
		internal_mul(b + mlen, n, a, mlen, scratch);
		internal_mod(a, mlen * 2, m, mlen, NULL, 0);
	    } else {
		BignumInt *t;
		t = a;
		a = b;
		b = t;
	    }
	    j--;
	}
	i++;
	j = BIGNUM_INT_BITS-1;
    }

    /* Fixup result in case the modulus was shifted */
    if (mshift) {
	for (i = mlen - 1; i < 2 * mlen - 1; i++)
	    a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
	a[2 * mlen - 1] = a[2 * mlen - 1] << mshift;
	internal_mod(a, mlen * 2, m, mlen, NULL, 0);
	for (i = 2 * mlen - 1; i >= mlen; i--)
	    a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
    }

    /* Copy result to buffer */
    result = newbn(mod[0]);
    for (i = 0; i < mlen; i++)
	result[result[0] - i] = a[i + mlen];
    while (result[0] > 1 && result[result[0]] == 0)
	result[0]--;

    /* Free temporary arrays */
    smemclr(a, 2 * mlen * sizeof(*a));
    sfree(a);
    smemclr(scratch, scratchlen * sizeof(*scratch));
    sfree(scratch);
    smemclr(b, 2 * mlen * sizeof(*b));
    sfree(b);
    smemclr(m, mlen * sizeof(*m));
    sfree(m);
    smemclr(n, mlen * sizeof(*n));
    sfree(n);

    freebn(base);

    return result;
}

/*
 * Compute (base ^ exp) % mod. Uses the Montgomery multiplication
 * technique where possible, falling back to modpow_simple otherwise.
 */
Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
{
    BignumInt *a, *b, *x, *n, *mninv, *scratch;
    int len, scratchlen, i, j;
    Bignum base, base2, r, rn, inv, result;

    /*
     * The most significant word of mod needs to be non-zero. It
     * should already be, but let's make sure.
     */
    assert(mod[mod[0]] != 0);

    /*
     * mod had better be odd, or we can't do Montgomery multiplication
     * using a power of two at all.
     */
    if (!(mod[1] & 1))
        return modpow_simple(base_in, exp, mod);

    /*
     * Make sure the base is smaller than the modulus, by reducing
     * it modulo the modulus if not.
     */
    base = bigmod(base_in, mod);

    /*
     * Compute the inverse of n mod r, for monty_reduce. (In fact we
     * want the inverse of _minus_ n mod r, but we'll sort that out
     * below.)
     */
    len = mod[0];
    r = bn_power_2(BIGNUM_INT_BITS * len);
    inv = modinv(mod, r);
    assert(inv); /* cannot fail, since mod is odd and r is a power of 2 */

    /*
     * Multiply the base by r mod n, to get it into Montgomery
     * representation.
     */
    base2 = modmul(base, r, mod);
    freebn(base);
    base = base2;

    rn = bigmod(r, mod);               /* r mod n, i.e. Montgomerified 1 */

    freebn(r);                         /* won't need this any more */

    /*
     * Set up internal arrays of the right lengths, in big-endian
     * format, containing the base, the modulus, and the modulus's
     * inverse.
     */
    n = snewn(len, BignumInt);
    for (j = 0; j < len; j++)
	n[len - 1 - j] = mod[j + 1];

    mninv = snewn(len, BignumInt);
    for (j = 0; j < len; j++)
	mninv[len - 1 - j] = (j < (int)inv[0] ? inv[j + 1] : 0);
    freebn(inv);         /* we don't need this copy of it any more */
    /* Now negate mninv mod r, so it's the inverse of -n rather than +n. */
    x = snewn(len, BignumInt);
    for (j = 0; j < len; j++)
        x[j] = 0;
    internal_sub(x, mninv, mninv, len);

    /* x = snewn(len, BignumInt); */ /* already done above */
    for (j = 0; j < len; j++)
	x[len - 1 - j] = (j < (int)base[0] ? base[j + 1] : 0);
    freebn(base);        /* we don't need this copy of it any more */

    a = snewn(2*len, BignumInt);
    b = snewn(2*len, BignumInt);
    for (j = 0; j < len; j++)
	a[2*len - 1 - j] = (j < (int)rn[0] ? rn[j + 1] : 0);
    freebn(rn);

    /* Scratch space for multiplies */
    scratchlen = 3*len + mul_compute_scratch(len);
    scratch = snewn(scratchlen, BignumInt);

    /* Skip leading zero bits of exp. */
    i = 0;
    j = BIGNUM_INT_BITS-1;
    while (i < (int)exp[0] && (exp[exp[0] - i] & ((BignumInt)1 << j)) == 0) {
	j--;
	if (j < 0) {
	    i++;
	    j = BIGNUM_INT_BITS-1;
	}
    }

    /* Main computation */
    while (i < (int)exp[0]) {
	while (j >= 0) {
	    internal_mul(a + len, a + len, b, len, scratch);
            monty_reduce(b, n, mninv, scratch, len);
	    if ((exp[exp[0] - i] & ((BignumInt)1 << j)) != 0) {
                internal_mul(b + len, x, a, len,  scratch);
                monty_reduce(a, n, mninv, scratch, len);
	    } else {
		BignumInt *t;
		t = a;
		a = b;
		b = t;
	    }
	    j--;
	}
	i++;
	j = BIGNUM_INT_BITS-1;
    }

    /*
     * Final monty_reduce to get back from the adjusted Montgomery
     * representation.
     */
    monty_reduce(a, n, mninv, scratch, len);

    /* Copy result to buffer */
    result = newbn(mod[0]);
    for (i = 0; i < len; i++)
	result[result[0] - i] = a[i + len];
    while (result[0] > 1 && result[result[0]] == 0)
	result[0]--;

    /* Free temporary arrays */
    smemclr(scratch, scratchlen * sizeof(*scratch));
    sfree(scratch);
    smemclr(a, 2 * len * sizeof(*a));
    sfree(a);
    smemclr(b, 2 * len * sizeof(*b));
    sfree(b);
    smemclr(mninv, len * sizeof(*mninv));
    sfree(mninv);
    smemclr(n, len * sizeof(*n));
    sfree(n);
    smemclr(x, len * sizeof(*x));
    sfree(x);

    return result;
}

/*
 * Compute (p * q) % mod.
 * The most significant word of mod MUST be non-zero.
 * We assume that the result array is the same size as the mod array.
 */
Bignum modmul(Bignum p, Bignum q, Bignum mod)
{
    BignumInt *a, *n, *m, *o, *scratch;
    int mshift, scratchlen;
    int pqlen, mlen, rlen, i, j;
    Bignum result;

    /*
     * The most significant word of mod needs to be non-zero. It
     * should already be, but let's make sure.
     */
    assert(mod[mod[0]] != 0);

    /* Allocate m of size mlen, copy mod to m */
    /* We use big endian internally */
    mlen = mod[0];
    m = snewn(mlen, BignumInt);
    for (j = 0; j < mlen; j++)
	m[j] = mod[mod[0] - j];

    /* Shift m left to make msb bit set */
    for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
	if ((m[0] << mshift) & BIGNUM_TOP_BIT)
	    break;
    if (mshift) {
	for (i = 0; i < mlen - 1; i++)
	    m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
	m[mlen - 1] = m[mlen - 1] << mshift;
    }

    pqlen = (p[0] > q[0] ? p[0] : q[0]);

    /*
     * Make sure that we're allowing enough space. The shifting below
     * will underflow the vectors we allocate if pqlen is too small.
     */
    if (2*pqlen <= mlen)
        pqlen = mlen/2 + 1;

    /* Allocate n of size pqlen, copy p to n */
    n = snewn(pqlen, BignumInt);
    i = pqlen - p[0];
    for (j = 0; j < i; j++)
	n[j] = 0;
    for (j = 0; j < (int)p[0]; j++)
	n[i + j] = p[p[0] - j];

    /* Allocate o of size pqlen, copy q to o */
    o = snewn(pqlen, BignumInt);
    i = pqlen - q[0];
    for (j = 0; j < i; j++)
	o[j] = 0;
    for (j = 0; j < (int)q[0]; j++)
	o[i + j] = q[q[0] - j];

    /* Allocate a of size 2*pqlen for result */
    a = snewn(2 * pqlen, BignumInt);

    /* Scratch space for multiplies */
    scratchlen = mul_compute_scratch(pqlen);
    scratch = snewn(scratchlen, BignumInt);

    /* Main computation */
    internal_mul(n, o, a, pqlen, scratch);
    internal_mod(a, pqlen * 2, m, mlen, NULL, 0);

    /* Fixup result in case the modulus was shifted */
    if (mshift) {
	for (i = 2 * pqlen - mlen - 1; i < 2 * pqlen - 1; i++)
	    a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
	a[2 * pqlen - 1] = a[2 * pqlen - 1] << mshift;
	internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
	for (i = 2 * pqlen - 1; i >= 2 * pqlen - mlen; i--)
	    a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
    }

    /* Copy result to buffer */
    rlen = (mlen < pqlen * 2 ? mlen : pqlen * 2);
    result = newbn(rlen);
    for (i = 0; i < rlen; i++)
	result[result[0] - i] = a[i + 2 * pqlen - rlen];
    while (result[0] > 1 && result[result[0]] == 0)
	result[0]--;

    /* Free temporary arrays */
    smemclr(scratch, scratchlen * sizeof(*scratch));
    sfree(scratch);
    smemclr(a, 2 * pqlen * sizeof(*a));
    sfree(a);
    smemclr(m, mlen * sizeof(*m));
    sfree(m);
    smemclr(n, pqlen * sizeof(*n));
    sfree(n);
    smemclr(o, pqlen * sizeof(*o));
    sfree(o);

    return result;
}

/*
 * Compute p % mod.
 * The most significant word of mod MUST be non-zero.
 * We assume that the result array is the same size as the mod array.
 * We optionally write out a quotient if `quotient' is non-NULL.
 * We can avoid writing out the result if `result' is NULL.
 */
static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient)
{
    BignumInt *n, *m;
    int mshift;
    int plen, mlen, i, j;

    /*
     * The most significant word of mod needs to be non-zero. It
     * should already be, but let's make sure.
     */
    assert(mod[mod[0]] != 0);

    /* Allocate m of size mlen, copy mod to m */
    /* We use big endian internally */
    mlen = mod[0];
    m = snewn(mlen, BignumInt);
    for (j = 0; j < mlen; j++)
	m[j] = mod[mod[0] - j];

    /* Shift m left to make msb bit set */
    for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
	if ((m[0] << mshift) & BIGNUM_TOP_BIT)
	    break;
    if (mshift) {
	for (i = 0; i < mlen - 1; i++)
	    m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
	m[mlen - 1] = m[mlen - 1] << mshift;
    }

    plen = p[0];
    /* Ensure plen > mlen */
    if (plen <= mlen)
	plen = mlen + 1;

    /* Allocate n of size plen, copy p to n */
    n = snewn(plen, BignumInt);
    for (j = 0; j < plen; j++)
	n[j] = 0;
    for (j = 1; j <= (int)p[0]; j++)
	n[plen - j] = p[j];

    /* Main computation */
    internal_mod(n, plen, m, mlen, quotient, mshift);

    /* Fixup result in case the modulus was shifted */
    if (mshift) {
	for (i = plen - mlen - 1; i < plen - 1; i++)
	    n[i] = (n[i] << mshift) | (n[i + 1] >> (BIGNUM_INT_BITS - mshift));
	n[plen - 1] = n[plen - 1] << mshift;
	internal_mod(n, plen, m, mlen, quotient, 0);
	for (i = plen - 1; i >= plen - mlen; i--)
	    n[i] = (n[i] >> mshift) | (n[i - 1] << (BIGNUM_INT_BITS - mshift));
    }

    /* Copy result to buffer */
    if (result) {
	for (i = 1; i <= (int)result[0]; i++) {
	    int j = plen - i;
	    result[i] = j >= 0 ? n[j] : 0;
	}
    }

    /* Free temporary arrays */
    smemclr(m, mlen * sizeof(*m));
    sfree(m);
    smemclr(n, plen * sizeof(*n));
    sfree(n);
}

/*
 * Decrement a number.
 */
void decbn(Bignum bn)
{
    int i = 1;
    while (i < (int)bn[0] && bn[i] == 0)
	bn[i++] = BIGNUM_INT_MASK;
    assert(i < (int)bn[0]);
    bn[i]--;
}

Bignum bignum_from_bytes(const unsigned char *data, int nbytes)
{
    Bignum result;
    int w, i;

    assert(nbytes >= 0 && nbytes < INT_MAX/8);

    w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */

    result = newbn(w);
    for (i = 1; i <= w; i++)
	result[i] = 0;
    for (i = nbytes; i--;) {
	unsigned char byte = *data++;
	result[1 + i / BIGNUM_INT_BYTES] |=
            (BignumInt)byte << (8*i % BIGNUM_INT_BITS);
    }

    while (result[0] > 1 && result[result[0]] == 0)
	result[0]--;
    return result;
}

/*
 * Read an SSH-1-format bignum from a data buffer. Return the number
 * of bytes consumed, or -1 if there wasn't enough data.
 */
int ssh1_read_bignum(const unsigned char *data, int len, Bignum * result)
{
    const unsigned char *p = data;
    int i;
    int w, b;

    if (len < 2)
	return -1;

    w = 0;
    for (i = 0; i < 2; i++)
	w = (w << 8) + *p++;
    b = (w + 7) / 8;		       /* bits -> bytes */

    if (len < b+2)
	return -1;

    if (!result)		       /* just return length */
	return b + 2;

    *result = bignum_from_bytes(p, b);

    return p + b - data;
}

/*
 * Return the bit count of a bignum, for SSH-1 encoding.
 */
int bignum_bitcount(Bignum bn)
{
    int bitcount = bn[0] * BIGNUM_INT_BITS - 1;
    while (bitcount >= 0
	   && (bn[bitcount / BIGNUM_INT_BITS + 1] >> (bitcount % BIGNUM_INT_BITS)) == 0) bitcount--;
    return bitcount + 1;
}

/*
 * Return the byte length of a bignum when SSH-1 encoded.
 */
int ssh1_bignum_length(Bignum bn)
{
    return 2 + (bignum_bitcount(bn) + 7) / 8;
}

/*
 * Return the byte length of a bignum when SSH-2 encoded.
 */
int ssh2_bignum_length(Bignum bn)
{
    return 4 + (bignum_bitcount(bn) + 8) / 8;
}

/*
 * Return a byte from a bignum; 0 is least significant, etc.
 */
int bignum_byte(Bignum bn, int i)
{
    if (i < 0 || i >= (int)(BIGNUM_INT_BYTES * bn[0]))
	return 0;		       /* beyond the end */
    else
	return (bn[i / BIGNUM_INT_BYTES + 1] >>
		((i % BIGNUM_INT_BYTES)*8)) & 0xFF;
}

/*
 * Return a bit from a bignum; 0 is least significant, etc.
 */
int bignum_bit(Bignum bn, int i)
{
    if (i < 0 || i >= (int)(BIGNUM_INT_BITS * bn[0]))
	return 0;		       /* beyond the end */
    else
	return (bn[i / BIGNUM_INT_BITS + 1] >> (i % BIGNUM_INT_BITS)) & 1;
}

/*
 * Set a bit in a bignum; 0 is least significant, etc.
 */
void bignum_set_bit(Bignum bn, int bitnum, int value)
{
    if (bitnum < 0 || bitnum >= (int)(BIGNUM_INT_BITS * bn[0])) {
        if (value) abort();		       /* beyond the end */
    } else {
	int v = bitnum / BIGNUM_INT_BITS + 1;
	BignumInt mask = (BignumInt)1 << (bitnum % BIGNUM_INT_BITS);
	if (value)
	    bn[v] |= mask;
	else
	    bn[v] &= ~mask;
    }
}

/*
 * Write a SSH-1-format bignum into a buffer. It is assumed the
 * buffer is big enough. Returns the number of bytes used.
 */
int ssh1_write_bignum(void *data, Bignum bn)
{
    unsigned char *p = data;
    int len = ssh1_bignum_length(bn);
    int i;
    int bitc = bignum_bitcount(bn);

    *p++ = (bitc >> 8) & 0xFF;
    *p++ = (bitc) & 0xFF;
    for (i = len - 2; i--;)
	*p++ = bignum_byte(bn, i);
    return len;
}

/*
 * Compare two bignums. Returns like strcmp.
 */
int bignum_cmp(Bignum a, Bignum b)
{
    int amax = a[0], bmax = b[0];
    int i;

    /* Annoyingly we have two representations of zero */
    if (amax == 1 && a[amax] == 0)
        amax = 0;
    if (bmax == 1 && b[bmax] == 0)
        bmax = 0;

    assert(amax == 0 || a[amax] != 0);
    assert(bmax == 0 || b[bmax] != 0);

    i = (amax > bmax ? amax : bmax);
    while (i) {
	BignumInt aval = (i > amax ? 0 : a[i]);
	BignumInt bval = (i > bmax ? 0 : b[i]);
	if (aval < bval)
	    return -1;
	if (aval > bval)
	    return +1;
	i--;
    }
    return 0;
}

/*
 * Right-shift one bignum to form another.
 */
Bignum bignum_rshift(Bignum a, int shift)
{
    Bignum ret;
    int i, shiftw, shiftb, shiftbb, bits;
    BignumInt ai, ai1;

    assert(shift >= 0);

    bits = bignum_bitcount(a) - shift;
    ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);

    if (ret) {
	shiftw = shift / BIGNUM_INT_BITS;
	shiftb = shift % BIGNUM_INT_BITS;
	shiftbb = BIGNUM_INT_BITS - shiftb;

	ai1 = a[shiftw + 1];
	for (i = 1; i <= (int)ret[0]; i++) {
	    ai = ai1;
	    ai1 = (i + shiftw + 1 <= (int)a[0] ? a[i + shiftw + 1] : 0);
	    ret[i] = ((ai >> shiftb) | (ai1 << shiftbb)) & BIGNUM_INT_MASK;
	}
    }

    return ret;
}

/*
 * Non-modular multiplication and addition.
 */
Bignum bigmuladd(Bignum a, Bignum b, Bignum addend)
{
    int alen = a[0], blen = b[0];
    int mlen = (alen > blen ? alen : blen);
    int rlen, i, maxspot;
    int wslen;
    BignumInt *workspace;
    Bignum ret;

    /* mlen space for a, mlen space for b, 2*mlen for result,
     * plus scratch space for multiplication */
    wslen = mlen * 4 + mul_compute_scratch(mlen);
    workspace = snewn(wslen, BignumInt);
    for (i = 0; i < mlen; i++) {
	workspace[0 * mlen + i] = (mlen - i <= (int)a[0] ? a[mlen - i] : 0);
	workspace[1 * mlen + i] = (mlen - i <= (int)b[0] ? b[mlen - i] : 0);
    }

    internal_mul(workspace + 0 * mlen, workspace + 1 * mlen,
		 workspace + 2 * mlen, mlen, workspace + 4 * mlen);

    /* now just copy the result back */
    rlen = alen + blen + 1;
    if (addend && rlen <= (int)addend[0])
	rlen = addend[0] + 1;
    ret = newbn(rlen);
    maxspot = 0;
    for (i = 1; i <= (int)ret[0]; i++) {
	ret[i] = (i <= 2 * mlen ? workspace[4 * mlen - i] : 0);
	if (ret[i] != 0)
	    maxspot = i;
    }
    ret[0] = maxspot;

    /* now add in the addend, if any */
    if (addend) {
	BignumDblInt carry = 0;
	for (i = 1; i <= rlen; i++) {
	    carry += (i <= (int)ret[0] ? ret[i] : 0);
	    carry += (i <= (int)addend[0] ? addend[i] : 0);
	    ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
	    carry >>= BIGNUM_INT_BITS;
	    if (ret[i] != 0 && i > maxspot)
		maxspot = i;
	}
    }
    ret[0] = maxspot;

    smemclr(workspace, wslen * sizeof(*workspace));
    sfree(workspace);
    return ret;
}

/*
 * Non-modular multiplication.
 */
Bignum bigmul(Bignum a, Bignum b)
{
    return bigmuladd(a, b, NULL);
}

/*
 * Simple addition.
 */
Bignum bigadd(Bignum a, Bignum b)
{
    int alen = a[0], blen = b[0];
    int rlen = (alen > blen ? alen : blen) + 1;
    int i, maxspot;
    Bignum ret;
    BignumDblInt carry;

    ret = newbn(rlen);

    carry = 0;
    maxspot = 0;
    for (i = 1; i <= rlen; i++) {
        carry += (i <= (int)a[0] ? a[i] : 0);
        carry += (i <= (int)b[0] ? b[i] : 0);
        ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
        carry >>= BIGNUM_INT_BITS;
        if (ret[i] != 0 && i > maxspot)
            maxspot = i;
    }
    ret[0] = maxspot;

    return ret;
}

/*
 * Subtraction. Returns a-b, or NULL if the result would come out
 * negative (recall that this entire bignum module only handles
 * positive numbers).
 */
Bignum bigsub(Bignum a, Bignum b)
{
    int alen = a[0], blen = b[0];
    int rlen = (alen > blen ? alen : blen);
    int i, maxspot;
    Bignum ret;
    BignumDblInt carry;

    ret = newbn(rlen);

    carry = 1;
    maxspot = 0;
    for (i = 1; i <= rlen; i++) {
        carry += (i <= (int)a[0] ? a[i] : 0);
        carry += (i <= (int)b[0] ? b[i] ^ BIGNUM_INT_MASK : BIGNUM_INT_MASK);
        ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
        carry >>= BIGNUM_INT_BITS;
        if (ret[i] != 0 && i > maxspot)
            maxspot = i;
    }
    ret[0] = maxspot;

    if (!carry) {
        freebn(ret);
        return NULL;
    }

    return ret;
}

/*
 * Create a bignum which is the bitmask covering another one. That
 * is, the smallest integer which is >= N and is also one less than
 * a power of two.
 */
Bignum bignum_bitmask(Bignum n)
{
    Bignum ret = copybn(n);
    int i;
    BignumInt j;

    i = ret[0];
    while (n[i] == 0 && i > 0)
	i--;
    if (i <= 0)
	return ret;		       /* input was zero */
    j = 1;
    while (j < n[i])
	j = 2 * j + 1;
    ret[i] = j;
    while (--i > 0)
	ret[i] = BIGNUM_INT_MASK;
    return ret;
}

/*
 * Convert a (max 32-bit) long into a bignum.
 */
Bignum bignum_from_long(unsigned long nn)
{
    Bignum ret;
    BignumDblInt n = nn;

    ret = newbn(3);
    ret[1] = (BignumInt)(n & BIGNUM_INT_MASK);
    ret[2] = (BignumInt)((n >> BIGNUM_INT_BITS) & BIGNUM_INT_MASK);
    ret[3] = 0;
    ret[0] = (ret[2]  ? 2 : 1);
    return ret;
}

/*
 * Add a long to a bignum.
 */
Bignum bignum_add_long(Bignum number, unsigned long addendx)
{
    Bignum ret = newbn(number[0] + 1);
    int i, maxspot = 0;
    BignumDblInt carry = 0, addend = addendx;

    for (i = 1; i <= (int)ret[0]; i++) {
	carry += addend & BIGNUM_INT_MASK;
	carry += (i <= (int)number[0] ? number[i] : 0);
	addend >>= BIGNUM_INT_BITS;
	ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
	carry >>= BIGNUM_INT_BITS;
	if (ret[i] != 0)
	    maxspot = i;
    }
    ret[0] = maxspot;
    return ret;
}

/*
 * Compute the residue of a bignum, modulo a (max 16-bit) short.
 */
unsigned short bignum_mod_short(Bignum number, unsigned short modulus)
{
    BignumDblInt mod, r;
    int i;

    r = 0;
    mod = modulus;
    for (i = number[0]; i > 0; i--)
	r = (r * (BIGNUM_TOP_BIT % mod) * 2 + number[i] % mod) % mod;
    return (unsigned short) r;
}

#ifdef DEBUG
void diagbn(char *prefix, Bignum md)
{
    int i, nibbles, morenibbles;
    static const char hex[] = "0123456789ABCDEF";

    debug(("%s0x", prefix ? prefix : ""));

    nibbles = (3 + bignum_bitcount(md)) / 4;
    if (nibbles < 1)
	nibbles = 1;
    morenibbles = 4 * md[0] - nibbles;
    for (i = 0; i < morenibbles; i++)
	debug(("-"));
    for (i = nibbles; i--;)
	debug(("%c",
	       hex[(bignum_byte(md, i / 2) >> (4 * (i % 2))) & 0xF]));

    if (prefix)
	debug(("\n"));
}
#endif

/*
 * Simple division.
 */
Bignum bigdiv(Bignum a, Bignum b)
{
    Bignum q = newbn(a[0]);
    bigdivmod(a, b, NULL, q);
    while (q[0] > 1 && q[q[0]] == 0)
        q[0]--;
    return q;
}

/*
 * Simple remainder.
 */
Bignum bigmod(Bignum a, Bignum b)
{
    Bignum r = newbn(b[0]);
    bigdivmod(a, b, r, NULL);
    while (r[0] > 1 && r[r[0]] == 0)
        r[0]--;
    return r;
}

/*
 * Greatest common divisor.
 */
Bignum biggcd(Bignum av, Bignum bv)
{
    Bignum a = copybn(av);
    Bignum b = copybn(bv);

    while (bignum_cmp(b, Zero) != 0) {
	Bignum t = newbn(b[0]);
	bigdivmod(a, b, t, NULL);
	while (t[0] > 1 && t[t[0]] == 0)
	    t[0]--;
	freebn(a);
	a = b;
	b = t;
    }

    freebn(b);
    return a;
}

/*
 * Modular inverse, using Euclid's extended algorithm.
 */
Bignum modinv(Bignum number, Bignum modulus)
{
    Bignum a = copybn(modulus);
    Bignum b = copybn(number);
    Bignum xp = copybn(Zero);
    Bignum x = copybn(One);
    int sign = +1;

    assert(number[number[0]] != 0);
    assert(modulus[modulus[0]] != 0);

    while (bignum_cmp(b, One) != 0) {
	Bignum t, q;

        if (bignum_cmp(b, Zero) == 0) {
            /*
             * Found a common factor between the inputs, so we cannot
             * return a modular inverse at all.
             */
            freebn(b);
            freebn(a);
            freebn(xp);
            freebn(x);
            return NULL;
        }

        t = newbn(b[0]);
	q = newbn(a[0]);
	bigdivmod(a, b, t, q);
	while (t[0] > 1 && t[t[0]] == 0)
	    t[0]--;
	while (q[0] > 1 && q[q[0]] == 0)
	    q[0]--;
	freebn(a);
	a = b;
	b = t;
	t = xp;
	xp = x;
	x = bigmuladd(q, xp, t);
	sign = -sign;
	freebn(t);
	freebn(q);
    }

    freebn(b);
    freebn(a);
    freebn(xp);

    /* now we know that sign * x == 1, and that x < modulus */
    if (sign < 0) {
	/* set a new x to be modulus - x */
	Bignum newx = newbn(modulus[0]);
	BignumInt carry = 0;
	int maxspot = 1;
	int i;

	for (i = 1; i <= (int)newx[0]; i++) {
	    BignumInt aword = (i <= (int)modulus[0] ? modulus[i] : 0);
	    BignumInt bword = (i <= (int)x[0] ? x[i] : 0);
	    newx[i] = aword - bword - carry;
	    bword = ~bword;
	    carry = carry ? (newx[i] >= bword) : (newx[i] > bword);
	    if (newx[i] != 0)
		maxspot = i;
	}
	newx[0] = maxspot;
	freebn(x);
	x = newx;
    }

    /* and return. */
    return x;
}

/*
 * Render a bignum into decimal. Return a malloced string holding
 * the decimal representation.
 */
char *bignum_decimal(Bignum x)
{
    int ndigits, ndigit;
    int i, iszero;
    BignumDblInt carry;
    char *ret;
    BignumInt *workspace;

    /*
     * First, estimate the number of digits. Since log(10)/log(2)
     * is just greater than 93/28 (the joys of continued fraction
     * approximations...) we know that for every 93 bits, we need
     * at most 28 digits. This will tell us how much to malloc.
     *
     * Formally: if x has i bits, that means x is strictly less
     * than 2^i. Since 2 is less than 10^(28/93), this is less than
     * 10^(28i/93). We need an integer power of ten, so we must
     * round up (rounding down might make it less than x again).
     * Therefore if we multiply the bit count by 28/93, rounding
     * up, we will have enough digits.
     *
     * i=0 (i.e., x=0) is an irritating special case.
     */
    i = bignum_bitcount(x);
    if (!i)
	ndigits = 1;		       /* x = 0 */
    else
	ndigits = (28 * i + 92) / 93;  /* multiply by 28/93 and round up */
    ndigits++;			       /* allow for trailing \0 */
    ret = snewn(ndigits, char);

    /*
     * Now allocate some workspace to hold the binary form as we
     * repeatedly divide it by ten. Initialise this to the
     * big-endian form of the number.
     */
    workspace = snewn(x[0], BignumInt);
    for (i = 0; i < (int)x[0]; i++)
	workspace[i] = x[x[0] - i];

    /*
     * Next, write the decimal number starting with the last digit.
     * We use ordinary short division, dividing 10 into the
     * workspace.
     */
    ndigit = ndigits - 1;
    ret[ndigit] = '\0';
    do {
	iszero = 1;
	carry = 0;
	for (i = 0; i < (int)x[0]; i++) {
	    carry = (carry << BIGNUM_INT_BITS) + workspace[i];
	    workspace[i] = (BignumInt) (carry / 10);
	    if (workspace[i])
		iszero = 0;
	    carry %= 10;
	}
	ret[--ndigit] = (char) (carry + '0');
    } while (!iszero);

    /*
     * There's a chance we've fallen short of the start of the
     * string. Correct if so.
     */
    if (ndigit > 0)
	memmove(ret, ret + ndigit, ndigits - ndigit);

    /*
     * Done.
     */
    smemclr(workspace, x[0] * sizeof(*workspace));
    sfree(workspace);
    return ret;
}

#ifdef TESTBN

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>

/*
 * gcc -Wall -g -O0 -DTESTBN -o testbn sshbn.c misc.c conf.c tree234.c unix/uxmisc.c -I. -I unix -I charset
 *
 * Then feed to this program's standard input the output of
 * testdata/bignum.py .
 */

void modalfatalbox(char *p, ...)
{
    va_list ap;
    fprintf(stderr, "FATAL ERROR: ");
    va_start(ap, p);
    vfprintf(stderr, p, ap);
    va_end(ap);
    fputc('\n', stderr);
    exit(1);
}

int random_byte(void)
{
    modalfatalbox("random_byte called in testbn");
    return 0;
}

#define fromxdigit(c) ( (c)>'9' ? ((c)&0xDF) - 'A' + 10 : (c) - '0' )

int main(int argc, char **argv)
{
    char *buf;
    int line = 0;
    int passes = 0, fails = 0;

    while ((buf = fgetline(stdin)) != NULL) {
        int maxlen = strlen(buf);
        unsigned char *data = snewn(maxlen, unsigned char);
        unsigned char *ptrs[5], *q;
        int ptrnum;
        char *bufp = buf;

        line++;

        q = data;
        ptrnum = 0;

        while (*bufp && !isspace((unsigned char)*bufp))
            bufp++;
        if (bufp)
            *bufp++ = '\0';

        while (*bufp) {
            char *start, *end;
            int i;

            while (*bufp && !isxdigit((unsigned char)*bufp))
                bufp++;
            start = bufp;

            if (!*bufp)
                break;

            while (*bufp && isxdigit((unsigned char)*bufp))
                bufp++;
            end = bufp;

            if (ptrnum >= lenof(ptrs))
                break;
            ptrs[ptrnum++] = q;
            
            for (i = -((end - start) & 1); i < end-start; i += 2) {
                unsigned char val = (i < 0 ? 0 : fromxdigit(start[i]));
                val = val * 16 + fromxdigit(start[i+1]);
                *q++ = val;
            }

            ptrs[ptrnum] = q;
        }

        if (!strcmp(buf, "mul")) {
            Bignum a, b, c, p;

            if (ptrnum != 3) {
                printf("%d: mul with %d parameters, expected 3\n", line, ptrnum);
                exit(1);
            }
            a = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
            b = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
            c = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
            p = bigmul(a, b);

            if (bignum_cmp(c, p) == 0) {
                passes++;
            } else {
                char *as = bignum_decimal(a);
                char *bs = bignum_decimal(b);
                char *cs = bignum_decimal(c);
                char *ps = bignum_decimal(p);
                
                printf("%d: fail: %s * %s gave %s expected %s\n",
                       line, as, bs, ps, cs);
                fails++;

                sfree(as);
                sfree(bs);
                sfree(cs);
                sfree(ps);
            }
            freebn(a);
            freebn(b);
            freebn(c);
            freebn(p);
        } else if (!strcmp(buf, "modmul")) {
            Bignum a, b, m, c, p;

            if (ptrnum != 4) {
                printf("%d: modmul with %d parameters, expected 4\n",
                       line, ptrnum);
                exit(1);
            }
            a = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
            b = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
            m = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
            c = bignum_from_bytes(ptrs[3], ptrs[4]-ptrs[3]);
            p = modmul(a, b, m);

            if (bignum_cmp(c, p) == 0) {
                passes++;
            } else {
                char *as = bignum_decimal(a);
                char *bs = bignum_decimal(b);
                char *ms = bignum_decimal(m);
                char *cs = bignum_decimal(c);
                char *ps = bignum_decimal(p);
                
                printf("%d: fail: %s * %s mod %s gave %s expected %s\n",
                       line, as, bs, ms, ps, cs);
                fails++;

                sfree(as);
                sfree(bs);
                sfree(ms);
                sfree(cs);
                sfree(ps);
            }
            freebn(a);
            freebn(b);
            freebn(m);
            freebn(c);
            freebn(p);
        } else if (!strcmp(buf, "pow")) {
            Bignum base, expt, modulus, expected, answer;

            if (ptrnum != 4) {
                printf("%d: mul with %d parameters, expected 4\n", line, ptrnum);
                exit(1);
            }

            base = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
            expt = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
            modulus = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
            expected = bignum_from_bytes(ptrs[3], ptrs[4]-ptrs[3]);
            answer = modpow(base, expt, modulus);

            if (bignum_cmp(expected, answer) == 0) {
                passes++;
            } else {
                char *as = bignum_decimal(base);
                char *bs = bignum_decimal(expt);
                char *cs = bignum_decimal(modulus);
                char *ds = bignum_decimal(answer);
                char *ps = bignum_decimal(expected);
                
                printf("%d: fail: %s ^ %s mod %s gave %s expected %s\n",
                       line, as, bs, cs, ds, ps);
                fails++;

                sfree(as);
                sfree(bs);
                sfree(cs);
                sfree(ds);
                sfree(ps);
            }
            freebn(base);
            freebn(expt);
            freebn(modulus);
            freebn(expected);
            freebn(answer);
        } else {
            printf("%d: unrecognised test keyword: '%s'\n", line, buf);
            exit(1);
        }

        sfree(buf);
        sfree(data);
    }

    printf("passed %d failed %d total %d\n", passes, fails, passes+fails);
    return fails != 0;
}

#endif