#include <bits/stdc++.h>
using namespace std;
set<long long> data;
void NOD(long long n) {
  long long res = 0;
  long long sqrtn = sqrt(n);
  for (long long i = 1; i < sqrtn; i++) {
    if (n % i == 0) {
      data.insert(i);
      data.insert(n / i);
    }
  }
  if (n % sqrtn == 0) {
    if (sqrtn * sqrtn == n) {
      data.insert(sqrtn);
    } else {
      data.insert(n / sqrtn);
      data.insert(sqrtn);
    }
  }
}
int main() {
  long long n, k;
  cin >> n >> k;
  NOD(n);
  set<long long>::iterator it;
  long long len = data.size();
  if (len >= k) {
    set<long long>::iterator it = data.begin();
    advance(it, k - 1);
    long long x = *it;
    cout << x << endl;
  } else {
    cout << -1 << endl;
  }
}

### 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;
set<long long> data;
void NOD(long long n) {
  long long res = 0;
  long long sqrtn = sqrt(n);
  for (long long i = 1; i < sqrtn; i++) {
    if (n % i == 0) {
      data.insert(i);
      data.insert(n / i);
    }
  }
  if (n % sqrtn == 0) {
    if (sqrtn * sqrtn == n) {
      data.insert(sqrtn);
    } else {
      data.insert(n / sqrtn);
      data.insert(sqrtn);
    }
  }
}
int main() {
  long long n, k;
  cin >> n >> k;
  NOD(n);
  set<long long>::iterator it;
  long long len = data.size();
  if (len >= k) {
    set<long long>::iterator it = data.begin();
    advance(it, k - 1);
    long long x = *it;
    cout << x << endl;
  } else {
    cout << -1 << endl;
  }
}

```