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;
const long long MAX = 1e5 + 5;
map<long long, long long> mp;
long long m[MAX];
long long a[MAX];
char s[MAX];
int main() {
  long long n;
  while (cin >> n) {
    scanf("%s", s);
    long long sz = strlen(s);
    a[0] = 0;
    for (long long i = 1; i <= sz; ++i) {
      a[i] = s[i - 1] - '0';
    }
    for (long long i = 2; i <= sz; ++i) a[i] += a[i - 1];
    memset(m, 0, sizeof(m));
    long long cnt = 1;
    mp.clear();
    for (long long i = 1; i <= (long long)sqrt(n * 1.0); ++i) {
      if (n % i == 0) {
        if (i == sqrt(n))
          mp[i] = cnt++;
        else {
          mp[i] = cnt++;
          mp[n / i] = cnt++;
        }
      }
    }
    long long c = 0, c0 = 0;
    for (long long i = 0; i <= sz; ++i) {
      for (long long j = i + 1; j <= sz; ++j) {
        long long tp = a[j] - a[i];
        if (tp != 0)
          c++;
        else
          c0++;
        if (mp.find(tp) != mp.end()) {
          m[mp[tp]]++;
        }
      }
    }
    long long tt = 0;
    if (n == 0) {
      cout << c0 * c0 + c0 * c * 2 << endl;
      continue;
    }
    long long ans = 0;
    for (long long i = 1; i <= (long long)sqrt(n * 1.0); ++i) {
      if (n % i == 0) {
        long long tt = n / i;
        if (mp.find(i) == mp.end() || mp.find(tt) == mp.end()) continue;
        if (i == sqrt(n))
          ans += m[mp[i]] * m[mp[i]];
        else
          ans += m[mp[i]] * m[mp[n / i]] * 2;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
