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:


////////////////////---------------------------SHUBHAM CHAUDHARI-------------------------------///////////////////////

import java.net.Inet4Address;
import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner in=new Scanner(System.in);
        long n=in.nextLong();
        int k=in.nextInt();
        Stack<Long>stack=new Stack<>();
        ArrayList<Long>list=new ArrayList<>();
        long res=-2;
        for(long i=1;i<Math.sqrt(n);i++)
        {
            if(n%i==0)
            {
                list.add(i);
                stack.push(i);
            }
        }
        if(n==((long)Math.sqrt(n)*(long)Math.sqrt(n))&&k>0)
        {
            list.add((long)Math.sqrt(n));
        }
        while (k>0&&stack.size()!=0)
        {
            list.add(n/stack.pop());
        }
        Collections.sort(list);
        if(k>list.size())
            System.out.println(-1);
        else
            System.out.println(list.get(k-1));

    }
}