#include <bits/stdc++.h>
using namespace std;
long long a[1000001], n, k, s = 0, d = 0;
int main() {
  cin >> n >> k;
  for (long long i = 1; i <= sqrt(n); i++)
    if (n % i == 0) {
      d += 2;
      if (i * i == n) {
        d--;
        a[s] = i;
        s++;
      } else {
        a[s] = i;
        a[s + 1] = n / i;
        s += 2;
      }
    }
  sort(a, a + s);
  if (d < k)
    cout << -1;
  else
    cout << a[k - 1];
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
long long a[1000001], n, k, s = 0, d = 0;
int main() {
  cin >> n >> k;
  for (long long i = 1; i <= sqrt(n); i++)
    if (n % i == 0) {
      d += 2;
      if (i * i == n) {
        d--;
        a[s] = i;
        s++;
      } else {
        a[s] = i;
        a[s + 1] = n / i;
        s += 2;
      }
    }
  sort(a, a + s);
  if (d < k)
    cout << -1;
  else
    cout << a[k - 1];
  return 0;
}

```