#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> factors;
  int cnt = 0;
  for (long long int i = 1; i * i <= n; i++)
    if (n % i == 0) {
      factors.push_back(i);
      cnt += 2;
      if (i * i == n) cnt--;
    }
  if (k <= cnt)
    if (k <= (cnt + 1) / 2)
      cout << factors[k - 1] << endl;
    else
      cout << n / factors[cnt - k] << endl;
  else
    cout << -1 << endl;
  return 0;
}

### 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 main() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> factors;
  int cnt = 0;
  for (long long int i = 1; i * i <= n; i++)
    if (n % i == 0) {
      factors.push_back(i);
      cnt += 2;
      if (i * i == n) cnt--;
    }
  if (k <= cnt)
    if (k <= (cnt + 1) / 2)
      cout << factors[k - 1] << endl;
    else
      cout << n / factors[cnt - k] << endl;
  else
    cout << -1 << endl;
  return 0;
}

```