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> divisor;
void divisores(long long N) {
  divisor.clear();
  for (long long div = 1, k; div * div <= N; div++) {
    if (N % div == 0) {
      divisor.push_back(div);
      k = N / div;
      if (k != div) divisor.push_back(k);
    }
  }
  sort(divisor.begin(), divisor.end());
}
int main() {
  long long n, k;
  cin >> n >> k;
  divisores(n);
  if (divisor.size() < k)
    cout << "-1\n";
  else
    cout << divisor[k - 1] << endl;
  return 0;
}