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[4005];
long long has[40000];
int main() {
  int i, j, k, q, a, n;
  while (~scanf("%d", &a)) {
    scanf("%s", &s);
    n = strlen(s);
    for (i = 0; i < n; i++) s[i] -= '0';
    memset(has, 0, sizeof has);
    for (i = 0; i < n; i++) {
      int su = 0;
      for (j = i; j < n; j++) {
        su += s[j];
        has[su]++;
      }
    }
    long long sum = 0;
    for (i = 0; i < n; i++) {
      int su = 0;
      for (j = i; j < n; j++) {
        su += s[j];
        if (su != 0 && a % su == 0 && a / su <= 9 * n)
          sum += has[a / su];
        else if (su == 0 && a == 0) {
          sum += n * (n - 1) / 2 + n;
        }
      }
    }
    printf("%I64d\n", sum);
  }
  return 0;
}
