#include <bits/stdc++.h>
using namespace std;
void solve(long long n, int k) {
  int j = 0;
  int i = 1;
  for (; i <= n / i; ++i) {
    if (n % i == 0) ++j;
    if (j == k) {
      cout << i << endl;
      return;
    }
  }
  while (i > 0 && i >= n / i) --i;
  for (; i > 0; --i) {
    if (n % i == 0) ++j;
    if (j == k) {
      cout << n / i << endl;
      return;
    }
  }
  cout << -1 << endl;
}
int main(void) {
  long long n;
  int k;
  cin >> n >> k;
  solve(n, k);
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
void solve(long long n, int k) {
  int j = 0;
  int i = 1;
  for (; i <= n / i; ++i) {
    if (n % i == 0) ++j;
    if (j == k) {
      cout << i << endl;
      return;
    }
  }
  while (i > 0 && i >= n / i) --i;
  for (; i > 0; --i) {
    if (n % i == 0) ++j;
    if (j == k) {
      cout << n / i << endl;
      return;
    }
  }
  cout << -1 << endl;
}
int main(void) {
  long long n;
  int k;
  cin >> n >> k;
  solve(n, k);
}

```