#include <bits/stdc++.h>
using namespace std;
long long int a[10000005];
int main() {
  long long int n, k;
  cin >> n >> k;
  int num = 1;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) a[num++] = i;
  }
  num--;
  if (a[num] * a[num] == n) {
    if (num * 2 - 1 < k)
      cout << "-1" << endl;
    else {
      if (num >= k)
        cout << a[k] << endl;
      else
        cout << n / a[2 * num - k] << endl;
    }
  } else {
    if (num * 2 < k) {
      cout << "-1" << endl;
      return 0;
    }
    if (num >= k)
      cout << a[k] << endl;
    else
      cout << n / a[2 * num - k + 1] << endl;
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int a[10000005];
int main() {
  long long int n, k;
  cin >> n >> k;
  int num = 1;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) a[num++] = i;
  }
  num--;
  if (a[num] * a[num] == n) {
    if (num * 2 - 1 < k)
      cout << "-1" << endl;
    else {
      if (num >= k)
        cout << a[k] << endl;
      else
        cout << n / a[2 * num - k] << endl;
    }
  } else {
    if (num * 2 < k) {
      cout << "-1" << endl;
      return 0;
    }
    if (num >= k)
      cout << a[k] << endl;
    else
      cout << n / a[2 * num - k + 1] << endl;
  }
  return 0;
}

```