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.

Solution (in JAVA):

Solution:
import java.util.*;
public class Main {
  public static void main(String[] args) {
    Scanner console = new Scanner(System.in);
    long n = console.nextLong();
    int k = console.nextInt();
    Vector<Long> small = new Vector<Long>();
    Vector<Long> big = new Vector<Long>();
    for (int i = 1; i <= Math.sqrt(n); i++) {
      if (n % i == 0) {
        small.add((long) i);
        if (n / i != i) {
          big.add(n/i);
        }
      }
    }
    Collections.reverse(big);
    if (k > small.size() + big.size()) {
      System.out.println(-1);
    } else if (k <= small.size()) {
        System.out.println(small.get(k-1));
    } else {
      System.out.println(big.get(k-1-small.size()));
    }
  }
}