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>
int mem[40000] = {0};
int main() {
  int a, b[4002], sum[4002], j, i;
  char s[4002];
  scanf("%d", &a);
  scanf("%s", s);
  i = 0;
  while (s[i] != '\0') {
    i++;
    b[i] = s[i - 1] - '0';
  }
  int k = i;
  sum[0] = 0;
  sum[1] = b[1];
  for (i = 2; i <= k; i++) sum[i] = sum[i - 1] + b[i];
  for (i = 0; i < k; i++) {
    for (j = i + 1; j <= k; j++) {
      mem[sum[j] - sum[i]]++;
    }
  }
  long long ans = 0, qq = 1;
  int q = sqrt(a);
  for (i = 1; i <= q; i++) {
    if (a % i == 0 && (a / i) <= sum[k]) ans += mem[i] * (mem[a / i] * qq);
  }
  if (a == 0) {
    ans += (mem[0] * qq) * (k * (k + 1) / 2);
  }
  ans = 2 * ans;
  if (q * q == a) ans -= mem[q] * (mem[q] * qq);
  printf("%I64d", ans);
  scanf(" ");
}
