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 c[4567];
long long d[4567];
map<long long, long long> m;
int main() {
  long long a;
  cin >> a;
  string s;
  cin >> s;
  if (a == 0) {
    long long h = 0;
    d[0] = 0;
    d[1] = s[0] - '0';
    for (long long i = 1; i < s.length(); i++) d[i + 1] = d[i] + s[i] - '0';
    for (long long i = 1; i < s.length() + 1; i++)
      for (long long j = i; j < s.length() + 1; j++)
        if (d[j] == d[i - 1]) h++;
    long long t = 0;
    for (long long i = 0; i < s.length(); i++)
      if (s[i] == '0') t++;
    long long ans = 0;
    for (long long i = 0; i < s.length(); i++)
      for (long long j = i; j < s.length(); j++)
        if (d[j + 1] - d[i] == 0)
          ans += (s.length() * (s.length() + 1)) / 2;
        else
          ans += h;
    cout << ans << endl;
    return 0;
  }
  d[0] = 0;
  d[1] = s[0] - '0';
  for (long long i = 1; i < s.length(); i++) {
    d[i + 1] = d[i] + s[i] - '0';
  }
  for (long long i = 1; i < s.length() + 1; i++)
    for (long long j = i; j < s.length() + 1; j++) m[d[j] - d[i - 1]]++;
  long long sch = 0;
  for (long long i = 1; i < s.length() + 1; i++)
    for (long long j = i; j < s.length() + 1; j++) {
      if (d[j] != d[i - 1] && a % (d[j] - d[i - 1]) == 0)
        sch += m[a / (d[j] - d[i - 1])];
    }
  cout << sch << endl;
  return 0;
}
