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

### 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() {
  long long num, a;
  cin >> num >> a;
  vector<long long> mv;
  long long square_root = (long long)sqrt(num) + 1;
  for (long long i = 1; i < square_root; i++) {
    if (num % i == 0 && i * i != num) {
      mv.push_back(i);
      mv.push_back(num / i);
    }
    if (num % i == 0 && i * i == num) mv.push_back(num / i);
  }
  if (mv.size() < a) {
    cout << -1 << endl;
    return 0;
  }
  sort(mv.begin(), mv.end());
  cout << mv[--a];
}

```