#include <bits/stdc++.h>
using namespace std;
unsigned long long n, k;
list<long long> l;
long long ans = -1;
inline void do_stuff() {
  long long t = 0;
  unsigned long long f = (unsigned long long)sqrt(n);
  for (int i = 1; i <= f; i++) {
    if (!(n % i)) {
      t++;
      l.push_back(i);
      if (t == k) {
        ans = i;
        return;
      }
    }
  }
  for (list<long long>::reverse_iterator it = l.rbegin(); it != l.rend();
       it++) {
    long long el = *it;
    if (n / el == el) continue;
    t++;
    if (t == k) {
      ans = n / el;
      return;
    }
  }
}
int main() {
  cin >> n >> k;
  do_stuff();
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
unsigned long long n, k;
list<long long> l;
long long ans = -1;
inline void do_stuff() {
  long long t = 0;
  unsigned long long f = (unsigned long long)sqrt(n);
  for (int i = 1; i <= f; i++) {
    if (!(n % i)) {
      t++;
      l.push_back(i);
      if (t == k) {
        ans = i;
        return;
      }
    }
  }
  for (list<long long>::reverse_iterator it = l.rbegin(); it != l.rend();
       it++) {
    long long el = *it;
    if (n / el == el) continue;
    t++;
    if (t == k) {
      ans = n / el;
      return;
    }
  }
}
int main() {
  cin >> n >> k;
  do_stuff();
  cout << ans << endl;
  return 0;
}

```