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, c = 0;
  vector<long long int> vec;
  scanf("%lld", &n);
  scanf("%lld", &k);
  for (long long int i = 1; i * i <= n; ++i) {
    if (n % i == 0) vec.push_back(i);
  }
  long long int last = vec.back(), div = vec.size() * 2;
  if (last * last == n) div--;
  if (k > div)
    printf("-1\n");
  else if (k <= vec.size()) {
    printf("%lld\n", vec[k - 1]);
  } else {
    printf("%lld\n", n / vec[div - k]);
  }
  return 0;
}