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 n, k;
int main() {
  while (cin >> n >> k) {
    long long blag = 0, s = sqrt(n), q;
    long long i, t = 0, ans = -1, p = 0;
    if (s * s == n) blag = 1;
    for (i = 1; i * i < n; i++) {
      if (n % i == 0) {
        t++;
        if (k == t) ans = i;
      }
    }
    if (ans == -1) {
      if (blag && k == t + 1)
        ans = sqrt(n);
      else {
        if (blag)
          q = 2 * t + 1;
        else
          q = 2 * t;
        if (k <= q && k > t) {
          k = q - k + 1;
          for (int i = 1; i * i < n; i++) {
            if (n % i == 0) p++;
            if (k == p) {
              ans = n / i;
              break;
            }
          }
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}