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>
int m, tot, ss[100010], w[100010];
int stop, sum[100010], n;
char s[100010];
long long all[100010];
void Fj(int a) {
  stop = (int)sqrt(a);
  for (int i = 1; i <= stop; i++)
    if (a % i == 0) {
      ss[++tot] = i;
      w[i] = tot;
      if (a / i != i) ss[++tot] = a / i;
    }
}
void get_all(int sum) {
  if (sum != 0 && m % sum != 0) return;
  if (sum <= stop)
    all[w[sum]]++;
  else
    all[w[m / sum] + 1]++;
}
void work() {
  long long ans = 0, all = n * (n + 1) / 2;
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + s[i] - '0';
    for (int j = 0; j < i; j++)
      if (sum[i] - sum[j] == 0) {
        ans += all * 2 - 1;
        all--;
      }
  }
  printf("%I64d\n", ans);
}
int main() {
  scanf("%d", &m);
  if (m != 0) Fj(m);
  scanf("%s", &s[1]);
  n = strlen(&s[1]);
  if (m == 0) {
    work();
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + s[i] - '0';
    for (int j = 0; j < i; j++) get_all(sum[i] - sum[j]);
  }
  long long ans = 0;
  for (int i = 1; i <= tot; i += 2) ans += all[i] * all[i + 1] * 2;
  if (tot % 2 == 1) ans += all[tot] * all[tot];
  printf("%I64d\n", ans);
  return 0;
}
