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;
#pragma comment(linker, "/STACK:256000000")
template <class T>
T sqr(T a) {
  return a * a;
}
int n, k;
int b, a;
long long m[4001];
long long cif[40001];
int kif[4001];
string s;
void init() {
  cin >> a;
  b = sqrt(a * 1.0);
  cin >> s;
  n = s.size();
  int q = 0;
  for (int i = 0; i < n; ++i) {
    kif[i] = s[i] - '0';
    if (!kif[i]) q = 1;
  }
  m[0] = kif[0];
  for (int i = 1; i < n; ++i) m[i] = m[i - 1] + kif[i];
  for (int i = 0; i < n; ++i)
    for (int j = i; j < n; ++j) cif[m[j] - m[i] + kif[i]]++;
}
void solve() {
  long long ot = 0;
  if (!a) {
    for (int i = 1; i <= 36000; ++i) {
      ot += cif[i] * cif[0];
    }
    ot *= 2;
    ot += cif[0] * cif[0];
  }
  for (int i = 1; i < 17; ++i) cerr << i << ' ' << cif[i] << endl;
  for (int i = 1; i <= min(36000, a); ++i) {
    if (a % i) continue;
    if (a / i > 40000) continue;
    ot += (cif[i] * cif[a / i]);
  }
  cout << ot;
}
int main() {
  init();
  solve();
  return 0;
}
