#include <bits/stdc++.h>
using namespace std;
long long n, k;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  vector<long long> factors;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      factors.push_back(i);
      if (n / i != i) {
        factors.push_back(n / i);
      }
    }
  }
  sort(factors.begin(), factors.end());
  if (factors.size() < k) {
    cout << -1;
  } else {
    cout << factors[k - 1];
  }
}

### Prompt

Create a solution in CPP for 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;
long long n, k;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  vector<long long> factors;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      factors.push_back(i);
      if (n / i != i) {
        factors.push_back(n / i);
      }
    }
  }
  sort(factors.begin(), factors.end());
  if (factors.size() < k) {
    cout << -1;
  } else {
    cout << factors[k - 1];
  }
}

```