#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, k;
  while (cin >> n >> k) {
    long long s = sqrt(n);
    long long cnt = 0;
    vector<long long> ans;
    for (__typeof(1ll) i = (1ll); i < (s + 1); i++) {
      if (n % i == 0) {
        ans.push_back(i);
        if (n / i != i) ans.push_back(n / i);
      }
    }
    sort(ans.begin(), ans.end());
    if (int((ans).size()) < k) {
      cout << -1 << '\n';
    } else {
      cout << ans[k - 1] << '\n';
    }
  }
  return 0;
}

### Prompt

Your task is to create 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::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, k;
  while (cin >> n >> k) {
    long long s = sqrt(n);
    long long cnt = 0;
    vector<long long> ans;
    for (__typeof(1ll) i = (1ll); i < (s + 1); i++) {
      if (n % i == 0) {
        ans.push_back(i);
        if (n / i != i) ans.push_back(n / i);
      }
    }
    sort(ans.begin(), ans.end());
    if (int((ans).size()) < k) {
      cout << -1 << '\n';
    } else {
      cout << ans[k - 1] << '\n';
    }
  }
  return 0;
}

```