#include <bits/stdc++.h>
using namespace std;
int num[10000010];
int main() {
  long long n, k, l, q = 0;
  cin >> n >> k;
  l = sqrt(n);
  for (int i = 1; i <= l; i++) {
    if (n % i == 0) {
      num[q++] = i;
    }
  }
  if (n % l == 0 && n / l == l)
    l = 2 * q - 1;
  else
    l = 2 * q;
  if (l < k)
    printf("-1\n");
  else if (q < k) {
    int t = k - q;
    if (l % 2 == 0) {
      printf("%I64d\n", (long long)(n / (num[q - t])));
    } else
      printf("%I64d\n", (long long)(n / (num[q - t - 1])));
  } else {
    printf("%d\n", num[k - 1]);
  }
  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;
int num[10000010];
int main() {
  long long n, k, l, q = 0;
  cin >> n >> k;
  l = sqrt(n);
  for (int i = 1; i <= l; i++) {
    if (n % i == 0) {
      num[q++] = i;
    }
  }
  if (n % l == 0 && n / l == l)
    l = 2 * q - 1;
  else
    l = 2 * q;
  if (l < k)
    printf("-1\n");
  else if (q < k) {
    int t = k - q;
    if (l % 2 == 0) {
      printf("%I64d\n", (long long)(n / (num[q - t])));
    } else
      printf("%I64d\n", (long long)(n / (num[q - t - 1])));
  } else {
    printf("%d\n", num[k - 1]);
  }
  return 0;
}

```