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;
template <class T>
void show(const vector<T> &a) {
  for (T x : a) cout << x << " ";
  cout << '\n';
}
const long long N = 3e5 + 50, oo = 1e18 + 100, mod = 1e9 + 7;
const long double eps = 1e-9, PI = 2 * acos(0.0);
vector<long long> vertices[N];
vector<long long> g[N];
vector<long long> visit(N, 0);
long long n, m, k;
long long cnt = 0;
const long long B = 4010;
long long pref[N];
long long col[N];
signed main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> m;
  string s;
  cin >> s;
  n = ((long long)(s.size()));
  for (long long i = 0; i < n; i++) {
    pref[i + 1] = pref[i] + s[i] - '0';
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = i; j <= n; j++) {
      col[pref[j] - pref[i - 1]]++;
    }
  }
  for (long long j = 1; j * j <= m; j++) {
    if (m % j == 0 && j <= pref[n] && m / j <= pref[n]) {
      cnt += col[j] * col[m / j];
      if (j != m / j) {
        cnt += col[j] * col[m / j];
      }
    }
  }
  if (!m) {
    cnt = col[0] * col[0];
    for (long long i = 1; i <= pref[n]; i++) {
      cnt += 2 * col[0] * col[i];
    }
  }
  cout << cnt;
  return 0;
}
