#include <bits/stdc++.h>
using namespace std;
int x;
long long d[10000001];
int main() {
  long long n;
  int k;
  cin >> n >> k;
  for (long long i = 1; i * i <= n; ++i) {
    if (n % i == 0) {
      d[x] = i;
      ++x;
      if (i * i != n) {
        d[x] = n / i;
        ++x;
      }
    }
  }
  if (x < k) {
    cout << -1;
    return 0;
  }
  sort(d, d + x + 1);
  cout << d[k];
}

### 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 x;
long long d[10000001];
int main() {
  long long n;
  int k;
  cin >> n >> k;
  for (long long i = 1; i * i <= n; ++i) {
    if (n % i == 0) {
      d[x] = i;
      ++x;
      if (i * i != n) {
        d[x] = n / i;
        ++x;
      }
    }
  }
  if (x < k) {
    cout << -1;
    return 0;
  }
  sort(d, d + x + 1);
  cout << d[k];
}

```