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.io.IOException;
import java.util.ArrayList;
import java.util.Scanner;
import java.util.TreeSet;


public class jakb {

    public static void main(String[] args) throws IOException {
        Scanner scanner = new Scanner(System.in);
        long n=scanner.nextLong();
        int k=scanner.nextInt();
        int l=0,m=0;
        TreeSet<Long>set=new TreeSet<>();
        for (long i = 1; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                set.add(i);
                set.add(n/i);
            }
        }
        if(k<set.size()+1){
            ArrayList<Long> a=new ArrayList<>(set);
            System.out.println(a.get(k-1));
        }
        else System.out.println("-1");





    }
}