#include <bits/stdc++.h>
using namespace std;
long long n, k;
int main() {
  cin >> n >> k;
  int d = 0;
  for (long long i = 1; i * i <= n; ++i) {
    if (n % i != 0) continue;
    if (d / 2 + 1 == k) {
      cout << i << endl;
      return 0;
    }
    if (i * i < n) {
      d += 2;
    } else {
      d++;
    }
  }
  if (k > d) {
    puts("-1");
    return 0;
  }
  int c = 0;
  for (long long i = 1; i * i <= n; ++i) {
    if (n % i != 0) continue;
    c++;
    if (d - c + 1 == k) {
      cout << n / i << endl;
      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 n, k;
int main() {
  cin >> n >> k;
  int d = 0;
  for (long long i = 1; i * i <= n; ++i) {
    if (n % i != 0) continue;
    if (d / 2 + 1 == k) {
      cout << i << endl;
      return 0;
    }
    if (i * i < n) {
      d += 2;
    } else {
      d++;
    }
  }
  if (k > d) {
    puts("-1");
    return 0;
  }
  int c = 0;
  for (long long i = 1; i * i <= n; ++i) {
    if (n % i != 0) continue;
    c++;
    if (d - c + 1 == k) {
      cout << n / i << endl;
      return 0;
    }
  }
}

```