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 a, n, sum[4002], w[1000000];
string s;
void work0() {
  sum[0] = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i - 1] == '0')
      sum[i] = sum[i - 1] + 1;
    else
      sum[i] = 0;
  }
  long long ans = 0;
  sum[n + 1] = 0;
  for (int i = 1; i <= n; i++)
    if (sum[i] > 0) {
      for (int j = 1; j <= n; j++)
        if (sum[j] > 0) {
          ans -= (long long)sum[i] * sum[j];
        }
    }
  for (int i = 1; i <= n; i++)
    if (sum[i] > 0) {
      ans += (long long)n * (n + 1) * sum[i];
    }
  cout << ans << endl;
}
void work1() {
  sum[0] = 0;
  memset(w, 0, sizeof(w));
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + s[i - 1] - '0';
  int hashId = 0;
  map<int, int> hash;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < i; j++) {
      int x = sum[i] - sum[j];
      if (x == 0 || a % x != 0) continue;
      int k;
      if (hash.find(x) == hash.end())
        hash[x] = k = hashId++;
      else
        k = hash[x];
      w[k]++;
    }
  }
  long long ans = 0;
  for (int i = 1; i * i <= a; i++)
    if (a % i == 0) {
      int x, y;
      if (hash.find(i) == hash.end()) continue;
      if (hash.find(a / i) == hash.end()) continue;
      x = hash[i];
      y = hash[a / i];
      ans += (long long)w[x] * w[y];
      if (i * i != a) ans += (long long)w[x] * w[y];
    }
  cout << ans << endl;
}
int main() {
  cin >> a;
  cin >> s;
  n = s.length();
  if (a == 0)
    work0();
  else
    work1();
  return 0;
}
