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 INF = 0x3fffffff;
const int MOD = 7;
const double EPS = 1e-7;
char s[100010];
int n, x, sum[100010];
int vis[100010];
int check(int y) {
  if (y == 0) {
    if (x == 0)
      return n * (n + 1) / 2;
    else
      return 0;
  }
  if ((x && x < y) || x % y) return 0;
  int k = x / y;
  if (k <= sum[n]) return vis[k];
  return 0;
}
int main() {
  scanf("%d", &x);
  scanf("%s", s);
  n = strlen(s);
  for (int i = 1; i <= n; i++) sum[i] = (s[i - 1] - '0') + sum[i - 1];
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) vis[sum[j] - sum[i - 1]]++;
  long long ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) ans += check(sum[j] - sum[i - 1]);
  printf("%I64d\n", ans);
  return 0;
}
