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 problemA {
    public static void main(String args[]){
        Scanner in = new Scanner(System.in);

        long n = in.nextLong();
        int k = in.nextInt();
        Set<Long> small = new TreeSet<>();
        for (int i = 1; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                small.add((long) i);
            }
        }
        Set<Long> big = new TreeSet<>();
        for (Long i : small) {
            big.add(n / i);
        }
        small.addAll(big);
        if (small.size() < k) {
            System.out.println(-1);
        } else {
            int i = 1;
            for (Long t : small) {
                if (i != k) {
                    i++;
                } else {
                    System.out.println(t);
                    break;
                }
            }
        }
    }

}