#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  long long int n, i, j, k, r;
  cin >> n >> k;
  r = sqrt(n) + 1;
  vector<long long int> div;
  j = 0;
  for (i = 1; i < r; i++) {
    if (n % i == 0) {
      j++;
      if (i != n / i) div.push_back(n / i);
    }
    if (j == k) {
      cout << i << endl;
      return 0;
    }
  }
  k = k - j;
  if (div.size() < k)
    cout << "-1" << endl;
  else {
    reverse(div.begin(), div.end());
    cout << div[k - 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() {
  ios_base::sync_with_stdio(false);
  long long int n, i, j, k, r;
  cin >> n >> k;
  r = sqrt(n) + 1;
  vector<long long int> div;
  j = 0;
  for (i = 1; i < r; i++) {
    if (n % i == 0) {
      j++;
      if (i != n / i) div.push_back(n / i);
    }
    if (j == k) {
      cout << i << endl;
      return 0;
    }
  }
  k = k - j;
  if (div.size() < k)
    cout << "-1" << endl;
  else {
    reverse(div.begin(), div.end());
    cout << div[k - 1] << endl;
  }
  return 0;
}

```