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 = 4010;
int cnt[36010];
long long ans;
int n;
int a[MaxN], r;
char buf[MaxN];
int main() {
  scanf("%d%s", &r, buf);
  n = strlen(buf);
  for (int i = 1; i <= n; i++) a[i] = a[i - 1] + buf[i - 1] - '0';
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < i; j++) cnt[a[i] - a[j]]++;
  if (r == 0)
    for (int i = 0; i <= 36000; i++) ans += (long long)cnt[0] * cnt[i];
  for (int i = 1; i <= 36000 && (r == 0 || i <= r); i++)
    if (cnt[i])
      if (r % i == 0 && r / i <= 36000) ans += (long long)cnt[i] * cnt[r / i];
  printf("%I64d\n", ans);
  return 0;
}
