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;
vector<long long> SUM;
map<long long, int> CNT;
long long SS[4010];
int main() {
  char s[5000];
  long long n;
  scanf("%I64d%s", &n, s);
  for (int i = 0; s[i]; ++i) {
    SS[i + 1] = SS[i] + s[i] - '0';
    for (int j = 0; j <= i; ++j) {
      SUM.push_back(SS[i + 1] - SS[j]);
      CNT[SS[i + 1] - SS[j]]++;
    }
  }
  int sz = SUM.size();
  long long ANS = 0;
  for (int i = 0; i < sz; ++i) {
    long long t;
    if (SUM[i])
      t = n / SUM[i];
    else
      t = 0;
    if (n == 0 && SUM[i] == 0)
      ANS += sz;
    else if (t * SUM[i] == n)
      ANS += CNT[t];
  }
  printf("%I64d\n", ANS);
  return 0;
}
