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 A {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        StringBuilder sb = new StringBuilder();

        long n = sc.nextLong();
        int k = sc.nextInt();

        ArrayList<Long> al = new ArrayList<>();

        long sqrt = (long) Math.sqrt(n);

        for (int i = 1; i <= sqrt; i++) {
            if (n % i == 0) {
                al.add((long) i);

                if (i != n / i) {
                    al.add(n / i);
                }

            }
        }

        Collections.sort(al);

        if (k > al.size()) {
            System.out.println(-1);
            return;
        }

        sb.append(al.get(k - 1)).append("\n");

        System.out.print(sb);
    }
}