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 Main{
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        long n = in.nextLong();
        long a = in.nextLong();
        
        int counter = 0;
        for (long i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                counter++;
                if (i * i != n){
                    counter++;
                }
            }
        }
        
        int count = 0;
        for (long i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                count++;
                if (count == a) {
                    System.out.println(i);
                    return;
                } else if (counter - count + 1 == a) {
                    System.out.println(n / i);
                    return;
                }
            }
        }
        
        System.out.println(-1);
    }
}