You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.

A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.

Input

The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).

Output

Print a single integer — the answer to a problem.

Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

10
12345


Output

6


Input

16
439873893693495623498263984765


Output

40
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T& RD(T& x) {
  char c;
  for (c = getchar(); c < '0'; c = getchar())
    ;
  x = c - '0';
  for (c = getchar(); '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  return x;
}
const int maxn = 4015;
long long cnt[40005];
char s[maxn];
int sum[maxn];
int main() {
  int k;
  while (~scanf("%d", &k)) {
    scanf("%s", s);
    int len = strlen(s);
    memset(cnt, 0, sizeof(cnt));
    memset(sum, 0, sizeof(sum));
    for (int i = 1; i <= len; i++) {
      sum[i] = sum[i - 1] + s[i - 1] - '0';
    }
    for (int i = 1; i <= len; i++) {
      for (int j = i; j <= len; j++) {
        cnt[sum[j] - sum[i - 1]]++;
      }
    }
    long long ans = 0;
    if (k == 0) {
      ans = 0;
      long long tem = 0;
      for (int i = 0; i <= sum[len]; i++) tem += cnt[i];
      ans = cnt[0] * tem * 2;
      ans -= cnt[0] * cnt[0];
    } else {
      for (int i = 1; i <= sum[len]; i++) {
        if (k % i == 0 && k / i <= sum[len]) {
          ans += cnt[i] * cnt[k / i];
        }
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
