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.*;
import java.util.*;
public class cfedq1
{
    public static void main(String args[])throws IOException
    {
        Scanner in=new Scanner(System.in);
        long n=in.nextLong();
        long k=in.nextLong();
        ArrayList<Long> arr=new ArrayList<Long>();
        long count=0;long i;
        for(i=1;i<=(long)Math.sqrt(n);i++)
        {
            if(n%i==0)
            {
                arr.add(i);
                if(n/i!=i)
                {
                    arr.add((long)(n/i));
                    count=count+2;
                }
                else
               count=count+1;
            }
            if(count>=(2*k))
            break;
        }
        long l=arr.size();
        if(l<k)
        System.out.println("-1");
        else
        {
        Collections.sort(arr);
       // for(int ii=0;ii<l;ii++)
       // System.out.print(arr.get(ii)+"\t");
        System.out.println(arr.get((int)k-1));
    }
        
    }
}