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;
long long num[5000];
long long divnum[37000];
int main() {
  long long n, sumi, sum = 0;
  long long p = 0;
  string str;
  cin >> n >> str;
  long long size = str.size();
  for (int i = 1; i <= size; i++) {
    num[i] = str[i - 1] - '0';
  }
  for (int i = 1; i <= size; i++) {
    sumi = 0;
    for (int j = i; j <= size; j++) {
      sumi += num[j];
      if (sumi != 0) {
        divnum[sumi]++;
      }
    }
  }
  if (n == 0) {
    for (int i = 0; i < size; ++i)
      if (str[i] == '0') {
        int j = i;
        while (str[j] == '0') {
          ++j;
        }
        p += (long long)(j - i) * (j - i + 1) / 2;
        i = j - 1;
      }
    long long ans =
        p * (size * (size + 1) / 2) + (size * (size + 1) / 2 - p) * p;
    cout << ans << endl;
    return 0;
  }
  for (long long i = 1; i * i <= n; i++) {
    if (i < 37000 && n / i < 37000) {
      if (n % i == 0) {
        sum += divnum[i] * divnum[n / i];
        if (i != (n / i)) {
          sum += divnum[i] * divnum[n / i];
        }
      }
    }
  }
  cout << sum;
}
