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;
vector<long long> divsL, divsR;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ;
  long long n, k;
  cin >> n >> k;
  for (long long i = 1; i * i <= n; i++)
    if (!(n % i)) {
      divsL.push_back(i);
      long long aux = n / i;
      if (aux != i) divsR.push_back(aux);
    }
  reverse((divsR).begin(), (divsR).end());
  if (k > ((int)divsL.size() + (int)divsR.size()))
    cout << "-1";
  else if (k <= (int)divsL.size())
    cout << divsL[k - 1];
  else
    cout << divsR[k - 1 - (int)divsL.size()];
  return 0;
}