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 n, k;
  cin >> n >> k;
  vector<long long> d;
  long long i = 1LL;
  while (i * i <= n) {
    if (n % i == 0) {
      d.push_back(i);
    }
    i++;
  }
  int ds = (int)d.size();
  for (int i = ds - 1; i >= 0; --i) {
    if (d[i] * d[i] != n) {
      d.push_back(n / d[i]);
    }
  }
  if ((int)d.size() < k) {
    printf("-1\n");
    return 0;
  }
  printf("%lld\n", d[k - 1]);
  return 0;
}