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.*;

public class Kdivisor {
    
    public static void main(String[] args) {
        
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        long k = sc.nextLong();

        if( k > n )
        {
            System.out.println("-1");
            return ;
        }
        
        Set<Long> set = new TreeSet<>();
        int temp = (int)Math.sqrt(n);
        for( long r=1 ; r <= temp ; r++ )
        {
            if (n % r == 0) {
                set.add(r);
                set.add(n/r);
            }
            
        }
        int x = 0;
        for( long i : set )
        {
            x++ ;
            if( x == k)
            {
                System.out.println(""+i);
                return;
            }
        }
        System.out.println("-1");
        
        
        
    }
    
}