#include <bits/stdc++.h>
using namespace std;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
long long n, k;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  vector<long long> div;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      div.push_back(i);
      if (n / i != i) {
        div.push_back(n / i);
      }
    }
  }
  sort(div.begin(), div.end());
  if ((int)div.size() < k) out(-1);
  out(div[k - 1]);
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
long long n, k;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  vector<long long> div;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      div.push_back(i);
      if (n / i != i) {
        div.push_back(n / i);
      }
    }
  }
  sort(div.begin(), div.end());
  if ((int)div.size() < k) out(-1);
  out(div[k - 1]);
  return 0;
}

```