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.ArrayList;
import java.util.Collections;
import java.util.Scanner;


public class Main {
    
    public static void main(String[] args) {
        
        Scanner input = new Scanner (System.in);
        
        long n = input.nextLong();
        int m = input.nextInt();
        
        ArrayList <Long> list = new ArrayList <Long> ();
        
        for(long i = 1 ; i <= Math.sqrt(n) ; i++)
        {
            if(n % i == 0)
            {
                list.add(i);
            
                if(i < Math.sqrt(n))
                    list.add(n/i);
            }
        }
        
        Collections.sort(list);
        
        if(m <= list.size())
            System.out.println(list.get(m-1));
        else
            System.out.println(-1);
        
    }
    
}