#include <bits/stdc++.h>
using namespace std;
#define N 1000005
#define P (int)(1e9 + 7)

char s[N];
int n, d, d2 = 1, pre[N], pw[N], inv[N], val[N], f[N][2], sum1[N], sum2[N], cnt2, cnt5;

int exgcd(int a, int b, int &x, int &y) {
	if (b == 0) {
		x = 1; y = 0;
		return a;
	}
	int d = exgcd(b, a % b, x, y);
	int t = x;
	x = y;
	y = t - (a / b) * y;
	return d;
}

int main() {
	scanf("%s%d", s + 1, &d);
	n = strlen(s + 1);
	// init cnt2 and cnt5
	for (; !(d & 1); ++cnt2, d >>= 1, d2 <<= 1);
	for (; !(d % 5); ++cnt5, d /= 5, d2 *= 5);
	// init hash
	pw[0] = 1;
	for (int i = 0, y; i <= n; i++) {
		if (i) pw[i] = pw[i - 1] * 10 % d,
			pre[i] = (pre[i - 1] * 10 + s[i] - '0') % d;
		exgcd(pw[i], d, inv[i], y);
		inv[i] = (inv[i] % d + d) % d;
		val[i] = 1LL * pre[i] * inv[i] % d;
	}
	// dp
	f[0][1] = 1;
	int all = 1, k = max({1, cnt2, cnt5});
	for (int i = 1; i <= n; i++) {
		// upd
		if (i >= k) {
			int j = i - k;
			sum1[val[j]] = (0LL + sum1[val[j]] + f[j][0] + f[j][1]) % P;
			sum2[val[j]] = (sum2[val[j]] + f[j][1]) % P;
		}
		// calc
		int res1 = 0, res2 = 0;
		if (i >= k) {
			int num = 0;
			for (int j = i - k + 1; j <= i; j++)
				num = (num * 10 + s[j] - '0') % d;
			if (!num)
				res1 = sum1[val[i]], res2 = sum2[val[i]];
		} int st = max(0, i - k + 1);
		int cur = 0, pw10 = 1;
		for (int j = i - 1; j >= st; --j) {
			cur = (cur + 1LL * (s[j + 1] - '0') * pw10 % d) % d;
			pw10 = pw10 * 10 % d;
			if (!cur) res1 = (0LL + res1 + f[j][0] + f[j][1]) % P,
				res2 = (res2 + f[j][1]) % P;
		}
		f[i][0] = (all - res2 + P) % P;
		f[i][1] = res1;
		all = (all + f[i][1]) % P;
	} printf("%d\n", (f[n][0] + f[n][1]) % P);
	return 0;
} 

样例一输出 $6$，没救了