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[4010];
long long int cnt[36002];
int main(void) {
  int i, j;
  long long int a, sum, ans;
  scanf("%I64d", &a);
  scanf("%s", str);
  memset(cnt, 0, sizeof(cnt));
  for (i = 0; str[i] != '\0'; i++) {
    sum = 0;
    for (j = i; str[j] != '\0'; j++) {
      sum += (str[j] - '0');
      ++cnt[sum];
    }
  }
  if (!a) {
    sum = 0;
    for (i = 1; i <= 36000; i++) sum += cnt[i];
    ans = (cnt[0] * sum) * 2 + cnt[0] * cnt[0];
  } else {
    ans = 0;
    for (i = 1; i <= a && i <= 36000; i++)
      if (!(a % i) && a / i <= 36000) ans += cnt[i] * cnt[a / i];
  }
  printf("%I64d\n", ans);
  return 0;
}
