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 a;
  cin >> a;
  string s;
  cin >> s;
  long long n = s.length() + 1;
  long long arr[n];
  arr[0] = 0;
  for (long long i = 1; i < n; i++) arr[i] = s[i - 1] - 48 + arr[i - 1];
  vector<long long> v;
  for (long long i = 0; i < n; i++)
    for (long long j = i + 1; j < n; j++) {
      long long x = (arr[j] - arr[i]);
      if (x > a && a != 0) break;
      if (a != 0 && x != 0 && (a % x == 0)) v.push_back(x);
      if (a == 0) v.push_back(x);
    }
  long long ans = 0;
  sort(v.begin(), v.end());
  int up0 = -1, low0 = -1;
  for (long long i = 0; i < v.size(); i++) {
    int up = 0, low = 0;
    if ((v[i] != 0 && a != 0) ||
        (v[i] != 0 && a == 0 && up0 == -1 && low0 == -1)) {
      up = upper_bound(v.begin(), v.end(), a / v[i]) - v.begin();
      low = lower_bound(v.begin(), v.end(), a / v[i]) - v.begin();
      if (v[up] == a / v[i]) up++;
      if (a == 0) up0 = up, low0 = low;
    }
    if (a != 0 && v[i] != 0 && a % v[i] == 0)
      ans = ans + up - low;
    else if (a == 0 && v[i] != 0)
      ans = ans + up0 - low0;
    else if (a == 0 && v[i] == 0)
      ans = ans + v.size();
  }
  cout << ans << endl;
  return 0;
}
