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, k;
  cin >> n >> k;
  set<long long> a = {1, n};
  for (long long d = 2; d <= min((long long)sqrt(n) + 10, n - 1); d++) {
    if (n % d == 0) {
      a.insert(d);
      a.insert(n / d);
    }
  }
  long long ans = -1;
  if (a.size() >= k) {
    auto it = next(a.begin(), k - 1);
    ans = *it;
  }
  cout << ans << endl;
}