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 cf762a {
    
    public static void main(String[] args) {
        
        Scanner input = new Scanner(System.in);
        long a = input.nextLong();
        int b = input.nextInt();
        
        int sqrt = (int) Math.sqrt(a);
        
        List<Long> d = new ArrayList<Long>();
        for (long i = 1; i <= sqrt; i++) {
            if (a % i == 0) {
                d.add(i);
                if (i != a / i) {
                    d.add(a / i);
                }
            }
        }
        Collections.sort(d);
        if (d.size() < b) {
            System.out.println(-1);
        } else {
            System.out.println(d.get(b - 1));
        }

        
    }
    
}