#include <bits/stdc++.h>
using namespace std;
int main() {
  vector<long long int> vc;
  long long int a, b;
  cin >> a >> b;
  for (long long int i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      vc.push_back(i);
      if (i != a / i) {
        vc.push_back(a / i);
      }
    }
  }
  sort(vc.begin(), vc.end());
  if (vc.size() < b) {
    cout << -1 << endl;
  } else {
    cout << vc[b - 1] << endl;
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
int main() {
  vector<long long int> vc;
  long long int a, b;
  cin >> a >> b;
  for (long long int i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      vc.push_back(i);
      if (i != a / i) {
        vc.push_back(a / i);
      }
    }
  }
  sort(vc.begin(), vc.end());
  if (vc.size() < b) {
    cout << -1 << endl;
  } else {
    cout << vc[b - 1] << endl;
  }
  return 0;
}

```