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;
int d[4010], n, m, x, y;
long long ans;
char st[4010];
int chk[40000];
int main() {
  scanf("%d%s", &n, st);
  m = (int)strlen(st);
  for (int i = 1; i <= m; i++) {
    if (i == 1)
      d[i] = st[i - 1] - '0';
    else
      d[i] = d[i - 1] + st[i - 1] - '0';
  }
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < i; j++) chk[d[i] - d[j]]++;
  if (n == 0)
    ans = (long long)m * (m + 1) * chk[0] - (long long)chk[0] * chk[0];
  else {
    for (int i = 1; i <= 36000; i++) {
      if (n % i == 0 && n / i <= 36000 && chk[n / i])
        ans += ((long long)chk[i] * chk[n / i]);
    }
  }
  printf("%lld", ans);
}
