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 str[4096];
int sum[4096];
int main() {
  int a;
  scanf("%d", &a);
  scanf("%s", str);
  int n = strlen(str);
  long long ret = 0;
  for (int i = 0; i < n; i++) {
    sum[i + 1] = sum[i] + str[i] - '0';
  }
  if (a == 0) {
    int c = 0;
    for (int j = 0; j < n; j++) {
      c += upper_bound(sum, sum + j + 1, sum[j + 1]) -
           lower_bound(sum, sum + j + 1, sum[j + 1]);
    }
    ret += (long long)(n * (n + 1) / 2) * (n * (n + 1) / 2) -
           (long long)(n * (n + 1) / 2 - c) * (n * (n + 1) / 2 - c);
  }
  for (int i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      int c = 0;
      for (int j = 0; j < n; j++) {
        c += upper_bound(sum, sum + j + 1, sum[j + 1] - i) -
             lower_bound(sum, sum + j + 1, sum[j + 1] - i);
      }
      int d = 0;
      for (int j = 0; j < n; j++)
        d += upper_bound(sum, sum + j + 1, sum[j + 1] - a / i) -
             lower_bound(sum, sum + j + 1, sum[j + 1] - a / i);
      ret += (long long)c * d;
      if (i * i != a) ret += (long long)c * d;
    }
  }
  printf("%I64d\n", ret);
}
