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 = 4005;
char s[maxn];
long long pre[maxn];
long long q[maxn * maxn];
long long li[maxn * maxn];
int le = 0;
int f(long long x) {
  int l = 0, r = le;
  while (l < r) {
    int mid = l + r >> 1;
    if (x <= li[mid])
      r = mid;
    else
      l = mid + 1;
  }
  if (li[l] == x)
    return l;
  else
    return maxn * maxn - 10;
}
int main() {
  long long a;
  cin >> a;
  cin >> s;
  long long len = strlen(s);
  pre[1] = s[0] - '0';
  for (int i = 2; i <= len; i++) pre[i] = pre[i - 1] + s[i - 1] - '0';
  int ll = 0;
  for (int i = 1; i <= len; i++)
    for (int j = i; j <= len; j++) {
      li[ll++] = pre[j] - pre[i - 1];
    }
  sort(li, li + ll);
  le = 0;
  for (int i = 1; i < ll; i++) {
    if (li[i] != li[le]) li[++le] = li[i];
  }
  for (int i = 1; i <= len; i++) {
    for (int j = i; j <= len; j++) {
      q[f(pre[j] - pre[i - 1])]++;
    }
  }
  long long ans = 0;
  if (a == 0)
    ans = q[f(0)] * len * (len + 1) - q[f(0)] * q[f(0)];
  else
    for (int i = 0; i <= le && li[i] < sqrt(a) + 0.5; i++) {
      if (li[i] == 0) continue;
      if (a % li[i] == 0) {
        if (li[i] == a / li[i])
          ans += q[i] * q[f(a / li[i])];
        else
          ans += 2LL * q[i] * q[f(a / li[i])];
      }
    }
  cout << ans << endl;
  return 0;
}
