#include <bits/stdc++.h>
int main() {
  long long int n, k, i, j, l = 0, flag = 0, array[100001];
  scanf("%lld %lld", &n, &k);
  long long int p = 0;
  for (i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      l++;
      if (n / i > sqrt(n)) {
        array[p] = n / i;
        p++;
      }
    }
    if (l == k) {
      flag = 1;
      break;
    }
  }
  if (flag == 0 && p - (k - l) >= 0) {
    printf("%lld\n", array[p - (k - l)]);
  } else if (flag == 1) {
    printf("%lld\n", i);
  } else
    printf("-1\n");
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
int main() {
  long long int n, k, i, j, l = 0, flag = 0, array[100001];
  scanf("%lld %lld", &n, &k);
  long long int p = 0;
  for (i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      l++;
      if (n / i > sqrt(n)) {
        array[p] = n / i;
        p++;
      }
    }
    if (l == k) {
      flag = 1;
      break;
    }
  }
  if (flag == 0 && p - (k - l) >= 0) {
    printf("%lld\n", array[p - (k - l)]);
  } else if (flag == 1) {
    printf("%lld\n", i);
  } else
    printf("-1\n");
  return 0;
}

```