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<pair<long long, long long> > v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> v1;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        v1.push_back(i);
        continue;
      }
      v1.push_back(i);
      v1.push_back(n / i);
    }
  }
  sort(v1.begin(), v1.end());
  if (k > v1.size()) {
    cout << "-1";
    return 0;
  }
  cout << v1[k - 1];
  return 0;
}