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;
int main() {
  long long int n, k, i;
  cin >> n >> k;
  long long int sq = (long long int)sqrt(n);
  int count = 0;
  int flag = 0;
  long long int re;
  for (i = 1; i <= sq; i++) {
    if (n % i == 0) {
      count++;
      if (k == count) {
        flag = 1;
        re = i;
        break;
      }
    }
  }
  if (flag == 0) {
    if ((sq * sq) == n)
      i = sq - 1;
    else
      i = sq;
    for (; i > 0; i--) {
      if (n % i == 0) {
        count++;
        if (k == count) {
          flag = 1;
          re = n / i;
          break;
        }
      }
    }
  }
  if (flag)
    cout << re << endl;
  else
    cout << -1 << endl;
  return 0;
}