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() {
  ios::sync_with_stdio(false);
  if (fopen("input.in", "r")) {
    freopen("input.in", "r", stdin);
    freopen("output.out", "w", stdout);
  }
  long long n, k;
  cin >> n >> k;
  long long a = sqrt(n);
  long long ct = 0;
  vector<long long> s;
  vector<long long> b;
  for (long long i = 1; i <= a; i++) {
    if (n % i == 0) {
      s.push_back(i);
      if (i * i != n) {
        b.push_back(n / i);
      }
    }
  }
  for (int i = b.size() - 1; i >= 0; i--) {
    s.push_back(b[i]);
  }
  if (s.size() < k) {
    cout << -1;
  } else
    cout << s[k - 1];
}