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 main() {
  long long int i, j, k, n;
  cin >> n;
  char a[1000000];
  cin >> a;
  if (a[1] == '\0') {
    if (n + '0' == a[0])
      cout << 1 << endl;
    else
      cout << 0 << endl;
    return 0;
  }
  long long pre[15000] = {0};
  pre[0] = a[i] - '0';
  for (i = 1; a[i] != '\0'; i++) {
    pre[i] = pre[i - 1] + (a[i] - '0');
  }
  int length = i;
  map<long long, long long int> hash;
  for (i = 0; i < length; i++) {
    for (j = i + 1; j < length; j++) {
      hash[pre[j] - pre[i]]++;
    }
  }
  for (i = 0; i < length; i++) hash[pre[i]]++;
  if (n == 0) {
    cout << hash[0] * (length * (length + 1) - hash[0]) << endl;
    return 0;
  }
  long long int ans = 0;
  for (i = 0; i < length; i++) {
    for (j = i + 1; j < length; j++) {
      long long sum = pre[j] - pre[i];
      if (sum != 0 && n % sum == 0) {
        long long int k = n / sum;
        if (hash[k] != 0) ans += hash[k];
      }
    }
  }
  for (i = 0; i < length; i++) {
    long long sum = pre[i];
    if (sum != 0 && n % sum == 0) {
      long long int k = n / sum;
      if (hash[k] != 0) ans += hash[k];
    }
  }
  cout << ans << endl;
  return 0;
}
