#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, k, i, s = 0, j, a, c, d, mas[100001] = {0}, b;
  cin >> a >> b;
  n = round(sqrt(a));
  for (i = 1; i <= n; ++i) {
    if (a % i == 0) {
      s++;
      mas[s] = i;
      s++;
      mas[s] = a / i;
    }
    if (i * i == a) s--;
  }
  sort(mas + 1, mas + s + 1);
  if (s < b)
    cout << -1;
  else
    cout << mas[b];
}

### 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;
int main() {
  long long n, k, i, s = 0, j, a, c, d, mas[100001] = {0}, b;
  cin >> a >> b;
  n = round(sqrt(a));
  for (i = 1; i <= n; ++i) {
    if (a % i == 0) {
      s++;
      mas[s] = i;
      s++;
      mas[s] = a / i;
    }
    if (i * i == a) s--;
  }
  sort(mas + 1, mas + s + 1);
  if (s < b)
    cout << -1;
  else
    cout << mas[b];
}

```