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;
  long long k;
  scanf("%lld%lld", &n, &k);
  vector<long long> v;
  long long l = 0, s = 0, temp = 0, i = 0;
  for (i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      l++;
      if (l == k) {
        temp = i;
        break;
      }
      if (n / i != i) {
        v.push_back(n / i);
      }
    }
  }
  s = v.size() - k + l;
  if (l != k && s >= 0) {
    printf("%lld", v[s]);
  } else if (s < 0)
    printf("-1\n");
  else
    printf("%lld", temp);
  return 0;
}