#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int test = 1;
  while (test--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v;
    for (long long i = 1; i * i <= n; i++) {
      if (n % i == 0) {
        v.push_back(n / i);
        k--;
        if (k == 0) {
          cout << i;
          return 0;
        }
      }
    }
    double x = sqrt(n);
    long long y = x;
    if ((double)y == x) k++;
    if (k > v.size()) {
      cout << "-1";
      return 0;
    }
    cout << v[v.size() - k];
  }
  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);
  cin.tie(0);
  cout.tie(0);
  ;
  int test = 1;
  while (test--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v;
    for (long long i = 1; i * i <= n; i++) {
      if (n % i == 0) {
        v.push_back(n / i);
        k--;
        if (k == 0) {
          cout << i;
          return 0;
        }
      }
    }
    double x = sqrt(n);
    long long y = x;
    if ((double)y == x) k++;
    if (k > v.size()) {
      cout << "-1";
      return 0;
    }
    cout << v[v.size() - k];
  }
  return 0;
}

```