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 int n;
  cin >> n;
  long long int k;
  cin >> k;
  set<long long int> st;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      st.insert(i);
      st.insert(n / i);
    }
  }
  st.insert(n / n);
  long long int cnt = 0;
  long long int r = 0;
  for (set<long long int>::iterator it = st.begin(); it != st.end(); it++) {
    cnt++;
    if (cnt == k) {
      r = *it;
      break;
    }
  }
  if (cnt < k) {
    cout << "-1" << endl;
  } else
    cout << r << endl;
}