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>
void quit();
using namespace std;
const long double PI = acos((double)-1.0);
const long double EPS = 1e-11;
double __t;
long long n, ts;
string s;
long long k[100100];
long long ans, knn;
int main() {
  cin >> n >> s;
  for (int l = 1; l <= s.size(); l++) {
    ts = 0;
    for (int j = 0; j < (int)l; ++j) ts += s[j] - '0';
    k[ts]++;
    if (ts > 0) knn++;
    for (int j = 0; j < (int)s.size() - l; ++j) {
      ts += s[j + l] - s[j];
      k[ts]++;
      if (ts > 0) knn++;
    }
  }
  if (n == 0) {
    cout << ((long long)(s.size() * (s.size() + 1)) / 2) *
                    ((long long)(s.size() * (s.size() + 1)) / 2) -
                knn * knn
         << endl;
    return 0;
  }
  for (int i = 1; i < 100100; i++)
    if (n % i == 0 && n / i < 100100) ans += k[i] * k[n / i];
  cout << ans;
  quit();
}
void quit() { exit(0); }
