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>
using min_queue = priority_queue<T, vector<T>, greater<T>>;
template <typename Args>
void kill(Args args) {
  cout << args << "\n";
  exit(0);
}
const double PI = acos(-1);
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const int N = 4040;
int val[N][N];
int arr[N];
string s;
int a;
int n;
map<int, int> mp;
void solve(int testnum) {
  cin >> a;
  cin >> s;
  n = (int)s.size();
  for (int i = 0; i < (int)s.size(); i++) {
    arr[i + 1] = s[i] - '0';
  }
  for (int i = 1; i <= n; i++) {
    arr[i] += arr[i - 1];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      val[i][j] = arr[j] - arr[i - 1];
      val[j][i] = arr[j] - arr[i - 1];
      mp[val[i][j]]++;
    }
  }
  long long ans = 0;
  if (!a) {
    ans = 1ll * (n * 1ll * (n + 1) * 1ll * mp[0]) - (mp[0] * 1ll * mp[0]);
    cout << ans << "\n";
    return;
  }
  for (int i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      ans += 1ll * (1 + (i * i != a)) * 1ll * mp[i] * 1ll * mp[a / i];
    }
  }
  cout << ans << "\n";
}
void precompute() {}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  int testcase = 1;
  clock_t zzzx = clock();
  precompute();
  for (int i = 1; i <= testcase; i++) {
    solve(i);
  }
  double elapsed_time = (double)(clock() - zzzx) / CLOCKS_PER_SEC;
  cerr << "elapsed_time"
       << " = " << (elapsed_time) << "\n";
  return 0;
}
