Problem: You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.

Divisor of n is any such natural number, that n can be divided by it without remainder.

Input

The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).

Output

If n has less than k divisors, output -1.

Otherwise, output the k-th smallest divisor of n.

Examples

Input

4 2


Output

2


Input

5 3


Output

-1


Input

12 5


Output

6

Note

In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.

In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long ans, res, p, q;
long long ara[1000005];
void solve() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      k--;
      if (k == 0) {
        cout << i << "\n";
        return;
      }
      if (i * i != n) {
        ara[res++] = n / i;
      }
    }
  }
  res--;
  for (long long i = res; i >= 0; i--) {
    k--;
    if (k == 0) {
      cout << ara[i] << "\n";
      return;
    }
  }
  cout << -1 << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tt = 1;
  while (tt--) solve();
}