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 main() {
  long long o = 0;
  long long a;
  string s;
  cin >> a >> s;
  long long n = s.size();
  vector<long long> sums(n);
  sums[0] = s[0] - '0';
  for (long long i = 1; i < n; i++) sums[i] = sums[i - 1] + s[i] - '0';
  vector<vector<long long> > sstr(n, vector<long long>(n));
  for (long long i = 0; i < n; i++) sstr[0][i] = sums[i];
  for (long long i = 0; i < n; i++) sstr[i][0] = s[i] - '0';
  vector<long long> count(400000, 0);
  for (long long i = 1; i < n; i++)
    for (long long j = 0; i + j < n; j++)
      sstr[i][j] = sstr[i - 1][j] + s[i + j] - s[i - 1];
  long long ans = 0;
  long long cwt0 = 0;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; i + j < n; j++) {
      count[sstr[i][j]]++;
      if (sstr[i][j] != 0) cwt0++;
      if (a == 0)
        ans += 2 * (count[0] + (sstr[i][j] == 0 ? cwt0 : 0));
      else if (sstr[i][j] && !(a % sstr[i][j]) && a / sstr[i][j] < 400000)
        ans += 2 * count[a / sstr[i][j]];
      if (sstr[i][j] * sstr[i][j] == a) ans--;
    }
  cout << ans << endl;
}
