#include <bits/stdc++.h>
using namespace std;
const int M = 10000005;
long long n, k, a[M], tot;
int main() {
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      a[++tot] = i;
      if (tot == k) {
        printf("%lld\n", i);
        break;
      }
    }
  }
  if (k > tot) {
    k -= tot;
    if (n / a[tot] == a[tot]) k++;
    if (tot + 1 > k)
      printf("%lld\n", n / a[tot + 1 - k]);
    else
      printf("-1\n");
  }
  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;
const int M = 10000005;
long long n, k, a[M], tot;
int main() {
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      a[++tot] = i;
      if (tot == k) {
        printf("%lld\n", i);
        break;
      }
    }
  }
  if (k > tot) {
    k -= tot;
    if (n / a[tot] == a[tot]) k++;
    if (tot + 1 > k)
      printf("%lld\n", n / a[tot + 1 - k]);
    else
      printf("-1\n");
  }
  return 0;
}

```