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 static java.lang.Math.sqrt;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;
//866421317361600
//26880

/**
 *
 * @author sharif
 */
public class CodeForces762A {

    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        long k = sc.nextLong();
        Set<Long> divisors = new TreeSet<>();

        for (long i = 1; i <= sqrt((double) n); i++) {
            if (n % i == 0) {
                divisors.add(i);
                divisors.add(n / i);
            }
        }
        if (divisors.size() >= k) {
            Iterator it = divisors.iterator();
            int i=0;
            while (it.hasNext()) {
               i++;
               if(i==k){
                   System.out.println(it.next());
                   break;
               }
               it.next();
            }
        } else {
            System.out.println("-1");
        }
    }
}