// PD ... np[x][y] nr de solutii cu restul y al unei configuratii x

#include <cstring>
#include <cstdio>
#include <algorithm>

int c[20], p;
long long n, np[1 << 18][20]; 

int main()
{
	freopen("ratphu.in", "r", stdin);
	freopen("ratphu.out", "w", stdout);
	scanf("%lld %lld", &n, &p);
	
	int s = 0;
	while (n)
	{
		c[s++] = n % 10;
		n /= 10;
	}
	std::reverse(c, c + s);
	
	np[0][0] = 1;
	for (int i = 0; i < (1 << s); ++i)
		for (int j = 0; j < p; ++j)
			if (np[i][j])
				for (int k = 0; k < s; ++k)
					if ((i & (1 << k)) == 0)
						np[i | (1 << k)][(j * 10 + c[k]) % p] += np[i][j];
	
	printf("%lld", np[(1 << s) - 1][0]);
}