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.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;
import java.util.Scanner;

/**
 * Created by abhijeet on 25/1/17.
 */
public class a {

    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        long n=in.nextLong();
        long k=in.nextLong();
        long count=0;
        long sqrt=(long)Math.sqrt(n);
        int flag=1;
        ArrayList<Long> a=new ArrayList<>();

        for(long i=1;i<=sqrt;i++)
        {
             if(n%i==0)
             {
                   if(i!=n/i)
                   {
                       a.add(i);
                       a.add(n/i);
                   }
                   else{
                       a.add(i);
                   }
             }

        }
        Collections.sort(a);
     //   System.out.print(a);
        if(k>a.size())
        {
            System.out.println("-1");
        }
        else{
            System.out.println(a.get((int)(k-1)));
        }




    }
}