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;
const long long MAX = 1e6 + 10, INF = 1e9 + 100;
void read(long long a[], long long n) {
  for (int(i) = 0; (i) < (n); (i)++) cin >> a[i];
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a;
  long long b = (long long)sqrt(n);
  for (unsigned long long i = 1; i <= b; i++) {
    if (n % i == 0 and i * i != n) {
      a.push_back(i);
      a.push_back(n / i);
    }
    if (n % i == 0 and i * i == n) {
      a.push_back(i);
    }
  }
  sort((a).begin(), (a).end());
  if (k > (int)(a).size()) {
    cout << -1 << "\n";
    return;
  }
  cout << a[k - 1] << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  solve();
  return 0;
}