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>
#pragma comment(linker, "/STACK:65777216")
using namespace std;
long long a;
string s;
long long sum[1000000];
long long ans = 0;
long long binpow(long long a1, long long b1) {
  if (b1 == 0) {
    return 1;
  }
  if (b1 % 2 == 1) {
    return binpow(a1, b1 - 1) * a1;
  } else {
    int b = binpow(a1, b1 / 2);
    return b * b;
  }
}
long long gcd(long long a1, long long b1) {
  if (b1 == 0)
    return a1;
  else
    return gcd(b1, a1 % b1);
}
int main() {
  cin >> a >> s;
  long long nu = 0;
  for (int i = 0; i < s.size(); i++) {
    long long sumt = 0;
    for (int j = i; j < s.size(); j++) {
      sumt += s[j] - '0';
      sum[sumt]++;
    }
  }
  long long z = 0;
  if (a == 0) {
    for (int i = 1; i <= 37000; i++) {
      nu += sum[0] * sum[i];
    }
    nu *= 2;
    z = (sum[0] * sum[0]);
    cout << nu + z;
    return 0;
  }
  for (long long i = 1; i <= 40000; i++) {
    if (a % i == 0 && sum[i] > 0 && a / i <= 40000 && sum[a / i] > 0) {
      ans += sum[i] * sum[a / i];
      if (i == a / i) {
        z++;
      }
    }
  }
  cout << ans - z / 2;
}
