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;
char s[4010];
int sum[4010];
int cnt[4010 * 9];
int main() {
  int a;
  scanf("%d%s", &a, s);
  int len = strlen(s);
  sum[0] = s[0] - '0';
  long long sol = 0;
  for (int i = 1; i < len; i++) sum[i] = (s[i] - '0') + sum[i - 1];
  for (int i = 0; i < len; i++)
    for (int j = i; j < len; j++) {
      int tmp = sum[j] - (i > 0 ? sum[i - 1] : 0);
      cnt[tmp]++;
    }
  for (int i = 0; i < len; i++)
    for (int j = i; j < len; j++) {
      int tmp = sum[j] - (i > 0 ? sum[i - 1] : 0);
      if (a == 0) {
        if (tmp != 0) {
          sol += (long long)cnt[0];
        } else
          sol += (long long)(len * (len + 1)) / 2;
      } else if (tmp != 0 && a % tmp == 0) {
        if (a / tmp < 4010 * 9) sol += (long long)cnt[a / tmp];
      }
    }
  printf("%lld\n", sol);
  return 0;
}
