#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n, k, j = 0, l = -1, m;
  cin >> n >> k;
  long long int arr[1000000];
  if (k > n) {
    cout << l;
    return 0;
  }
  for (long long int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0 && i * i != n) {
      arr[j++] = i;
      arr[j++] = n / i;
    }
    if (i * i == n) arr[j++] = i;
  }
  arr[j++] = 1;
  if (n != 1) arr[j++] = n;
  arr[j] = 0;
  sort(arr, arr + j + 1);
  if (j < k)
    cout << l;
  else
    cout << arr[k];
  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;
int main() {
  long long int n, k, j = 0, l = -1, m;
  cin >> n >> k;
  long long int arr[1000000];
  if (k > n) {
    cout << l;
    return 0;
  }
  for (long long int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0 && i * i != n) {
      arr[j++] = i;
      arr[j++] = n / i;
    }
    if (i * i == n) arr[j++] = i;
  }
  arr[j++] = 1;
  if (n != 1) arr[j++] = n;
  arr[j] = 0;
  sort(arr, arr + j + 1);
  if (j < k)
    cout << l;
  else
    cout << arr[k];
  return 0;
}

```