#include <bits/stdc++.h>
using namespace std;
vector<long long> v;
int main() {
  long long a, b;
  int alert = 0;
  cin >> a >> b;
  for (long long i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      v.push_back(i);
      if (i * i == a) alert = 1;
    }
  }
  int k = v.size();
  b--;
  if (b < v.size()) {
    cout << v[b];
  } else if (b + alert >= 2 * v.size()) {
    cout << -1;
  } else {
    cout << a / v[2 * v.size() - b - alert - 1];
  }
  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;
vector<long long> v;
int main() {
  long long a, b;
  int alert = 0;
  cin >> a >> b;
  for (long long i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      v.push_back(i);
      if (i * i == a) alert = 1;
    }
  }
  int k = v.size();
  b--;
  if (b < v.size()) {
    cout << v[b];
  } else if (b + alert >= 2 * v.size()) {
    cout << -1;
  } else {
    cout << a / v[2 * v.size() - b - alert - 1];
  }
  return 0;
}

```