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 ans = 0, a, mtx[4001], sums[4001 * 9] = {};
  string s;
  cin >> a >> s;
  int l = s.length();
  for (int i = 0; i < l; i++)
    if (i)
      mtx[i] = s[i] - 48 + mtx[i - 1];
    else
      mtx[i] = s[i] - 48;
  for (int i = 0; i < l; i++)
    for (int j = i; j < l; j++)
      if (i)
        sums[mtx[j] - mtx[i - 1]]++;
      else
        sums[mtx[j]]++;
  for (int i = 0; i < l; i++)
    for (int j = i; j < l; j++)
      if (i) {
        if (a == 0 && mtx[j] - mtx[i - 1] == 0)
          ans += l * (l + 1) / 2;
        else if ((mtx[j] - mtx[i - 1]))
          if (a % (mtx[j] - mtx[i - 1]) == 0)
            if ((a / (mtx[j] - mtx[i - 1])) <= 4000 * 9)
              ans += sums[a / (mtx[j] - mtx[i - 1])];
      } else {
        if (a == 0 && mtx[j] == 0)
          ans += l * (l + 1) / 2;
        else if (mtx[j])
          if (a % mtx[j] == 0)
            if ((a / mtx[j]) <= 4000 * 9) ans += sums[a / mtx[j]];
      }
  cout << ans << endl;
  return 0;
}
