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 int gcd(long long int a, long long int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans % m;
}
int main() {
  long long int a, x = 0, z = 0, ans = 0, n, c = 0, d = 0, ss = 0;
  cin >> a;
  string s, r;
  cin >> s;
  n = s.size();
  map<long long int, long long int> m;
  vector<long long int> v;
  if (a == 0) {
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == '0') {
        c = 0;
        while (s[i] == '0' && i < n) {
          i++;
          c++;
        }
        ss += (c * (c + 1)) / 2;
      }
    }
    ans += ss * (n * (n + 1));
    ans -= (ss * ss);
  } else {
    for (int i = 0; i < s.size(); i++) {
      x = 0;
      for (int j = i; j < s.size(); j++) {
        m[x + s[j] - '0']++;
        x += (s[j] - '0');
      }
    }
    for (int i = 1; i * i <= a; i++) {
      if (a % i == 0) {
        if (i * i == a)
          ans += (m[a / i] * m[i]);
        else
          ans += (2 * m[a / i] * m[i]);
      }
    }
  }
  cout << ans;
}
