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 int MAXN = 4005;
char c[MAXN];
int sum[MAXN][MAXN];
map<int, int> cnt;
map<int, int>::iterator it;
void solve() {
  int n, i, j, a;
  scanf("%d%s", &a, &c);
  n = strlen(c);
  for (i = 0; i < n; i++) {
    c[i] -= '0';
  }
  for (i = n - 1; i >= 0; i--) {
    sum[i][i] = c[i];
    for (j = i + 1; j < n; j++) {
      sum[i][j] = sum[i][j - 1] + c[j];
    }
  }
  long long ans = 0, tmp, tmp1, cnt1;
  if (a == 0) {
    cnt1 = 0;
    for (i = 0; i < n; i++) {
      for (j = i; j < n; j++) {
        if (sum[i][j] == 0) {
          cnt1++;
          ans += n * (n + 1);
        }
      }
    }
    ans -= cnt1 * cnt1;
    printf("%I64d", ans);
    return;
  }
  for (i = 0; i < n; i++) {
    for (j = i; j < n; j++) {
      if (sum[i][j] != 0 && a % sum[i][j] == 0) {
        cnt[sum[i][j]]++;
      }
    }
  }
  for (it = cnt.begin(); it != cnt.end(); it++) {
    tmp = it->first;
    tmp1 = (cnt.find(a / tmp) == cnt.end() ? 0 : cnt[a / tmp]);
    ans += (long long)cnt[tmp] * tmp1;
  }
  printf("%I64d", ans);
}
int main() {
  int t;
  t = 1;
  while (t) {
    t--;
    solve();
  }
}
