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

### Prompt

In Cpp, your task is to solve 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 int k, n;
  while (cin >> k >> n) {
    set<long long> num;
    for (int i = 1; i <= sqrt(k); i++) {
      if (k % i == 0) {
        num.insert(i);
        num.insert(k / i);
      }
    }
    if (n > num.size())
      cout << -1 << endl;
    else {
      set<long long>::iterator it = num.begin();
      while (n--) it++;
      it--;
      cout << *it << endl;
    }
  }
  return 0;
}

```