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;
int dirx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int diry[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
const int N = 4e3 + 5;
int a, p[N];
long long ans;
char s[N];
map<int, long long> m;
int main() {
  scanf("%d%s", &a, s);
  long long n = strlen(s);
  for (int i = 0; s[i]; i++) p[i + 1] = p[i] + (s[i] - '0');
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < i; j++) m[p[i] - p[j]]++;
  if (a == 0) ans = m[0] * (n + 1) * n - m[0] * m[0];
  int i = 1;
  for (i = 1; i * i < a; i++)
    if (a % i == 0) ans += 2 * m[i] * m[a / i];
  if (i * i == a) ans += m[i] * m[i];
  printf("%lld\n", ans);
  return 0;
}
