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;
void solve() {
  long long n;
  cin >> n;
  ;
  long long m;
  cin >> m;
  ;
  set<long long> s;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      s.insert(i);
      s.insert(n / i);
    }
  }
  if (s.size() < m) {
    cout << -1;
    return;
  }
  auto it = s.begin();
  for (long long i = 1; it != s.end() && i < m; it++, i++)
    ;
  cout << *(it);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}