#include <bits/stdc++.h>
using namespace std;
int main() {
  long long num, answer;
  unsigned int k, max_number_of_divisors;
  scanf("%I64d %d", &num, &k);
  vector<int> divisors;
  for (int i = 1; i * 1LL * i <= num; i++)
    if (num % i == 0) divisors.push_back(i);
  max_number_of_divisors = 2 * divisors.size();
  if (divisors.back() * 1LL * divisors.back() == num) max_number_of_divisors--;
  if (k > max_number_of_divisors) {
    answer = -1;
  } else {
    if (k <= divisors.size())
      answer = divisors[k - 1];
    else
      answer = num / divisors[max_number_of_divisors - k];
  }
  printf("%I64d", answer);
  return 0;
}

### Prompt

Generate a cpp solution 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;
int main() {
  long long num, answer;
  unsigned int k, max_number_of_divisors;
  scanf("%I64d %d", &num, &k);
  vector<int> divisors;
  for (int i = 1; i * 1LL * i <= num; i++)
    if (num % i == 0) divisors.push_back(i);
  max_number_of_divisors = 2 * divisors.size();
  if (divisors.back() * 1LL * divisors.back() == num) max_number_of_divisors--;
  if (k > max_number_of_divisors) {
    answer = -1;
  } else {
    if (k <= divisors.size())
      answer = divisors[k - 1];
    else
      answer = num / divisors[max_number_of_divisors - k];
  }
  printf("%I64d", answer);
  return 0;
}

```