#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long i, j, k;
  long long n, sum;
  cin >> n >> k;
  set<long long> v;
  for (i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      v.insert(i);
      v.insert(n / i);
    }
  }
  if (v.size() < k)
    cout << "-1";
  else {
    set<long long>::iterator p = v.begin();
    j = 1;
    while (1) {
      if (j == k) {
        cout << *p;
        break;
      }
      p++;
      j++;
    }
  }
  return 0;
}

### Prompt

Please formulate 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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long i, j, k;
  long long n, sum;
  cin >> n >> k;
  set<long long> v;
  for (i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      v.insert(i);
      v.insert(n / i);
    }
  }
  if (v.size() < k)
    cout << "-1";
  else {
    set<long long>::iterator p = v.begin();
    j = 1;
    while (1) {
      if (j == k) {
        cout << *p;
        break;
      }
      p++;
      j++;
    }
  }
  return 0;
}

```