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.*;
import java.math.*;
public class nit{
    
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        long n=sc.nextLong();
        int k=sc.nextInt();boolean flag=true;
        HashSet<Long> l=new HashSet<>();
        long m=1;
        while(m<=Math.floor(Math.sqrt(n))){
            if(n%m==0){
                long g=n/m;
               
                l.add(g);
                l.add(m);      
            }
            m++;
            
        }
        ArrayList<Long> lm=new ArrayList<>(l);
        Collections.sort(lm);
        //System.out.println(lm);
        if(lm.size()!=0&&lm.size()>=(k)){
           System.out.println(lm.get(k-1));
           flag=false;
        }
                    
        if(flag)
            System.out.println("-1");
    }
}