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

### Prompt

Develop a solution in CPP 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;
int main() {
  long long n, k;
  cin >> n >> k;
  vector<long long> vec;
  long long temp = 0;
  long long i = 0;
  for (i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      temp++;
      vec.push_back(i);
    }
    if (temp == k) {
      cout << vec[vec.size() - 1] << endl;
      return 0;
    }
  }
  i--;
  if (i == sqrt(n)) {
    if (k > (2 * (vec.size() - 1) + 1)) {
      cout << -1 << endl;
      return 0;
    } else {
      long long rem = k - vec.size();
      cout << n / vec[vec.size() - rem - 1];
    }
  } else {
    if (k > (2 * vec.size())) {
      cout << -1 << endl;
      return 0;
    } else {
      long long rem = k - vec.size();
      cout << n / vec[vec.size() - rem] << endl;
    }
  }
}

```