#include <bits/stdc++.h>
using namespace std;
int main() {
  long long num, k;
  cin >> num >> k;
  k--;
  vector<long long> vs, vl;
  int sq = sqrt(num + 1);
  for (int i = 1; i <= sq; i++) {
    if (num % i == 0) {
      vs.push_back(i);
      if (num / i != i) vl.push_back(num / i);
    }
  }
  if (k + 1 > (vs.size() + vl.size())) {
    cout << "-1" << endl;
    return 0;
  }
  if (num == 1 and k > 0) {
    cout << "-1" << endl;
    return 0;
  }
  if (vs.size() > k)
    cout << vs[k] << endl;
  else
    cout << vl[vl.size() + vs.size() - 1 - k] << endl;
  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() {
  long long num, k;
  cin >> num >> k;
  k--;
  vector<long long> vs, vl;
  int sq = sqrt(num + 1);
  for (int i = 1; i <= sq; i++) {
    if (num % i == 0) {
      vs.push_back(i);
      if (num / i != i) vl.push_back(num / i);
    }
  }
  if (k + 1 > (vs.size() + vl.size())) {
    cout << "-1" << endl;
    return 0;
  }
  if (num == 1 and k > 0) {
    cout << "-1" << endl;
    return 0;
  }
  if (vs.size() > k)
    cout << vs[k] << endl;
  else
    cout << vl[vl.size() + vs.size() - 1 - k] << endl;
  return 0;
}

```