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 N = 4005;
const int M = 36005;
char str[N];
int sum[N];
long long num[M];
int main() {
  int a;
  scanf("%d%s", &a, str);
  int l = strlen(str);
  for (int i = 0; i < l; ++i) str[i] -= '0';
  sum[0] = 0;
  for (int i = 1; i <= l; ++i) sum[i] = str[i - 1] + sum[i - 1];
  memset(num, 0, sizeof(num));
  long long cnt = 0;
  for (int i = 0; i < l; ++i)
    for (int j = i + 1; j <= l; ++j) {
      int t = sum[j] - sum[i];
      ++num[t];
      if (t * t == a) ++cnt;
    }
  long long ans = 0;
  if (a == 0)
    for (int i = 1; i < M; ++i) ans += num[0] * num[i];
  for (int i = 1; i * i < a; ++i)
    if (a % i == 0) {
      int j = a / i;
      if (j < M) ans += num[i] * num[j];
    }
  printf("%I64d", ans * 2 + cnt * cnt);
}
