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.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

/**
 * Created by llee on 27/01/17.
 */
public class KthDivisor {
    static List<Long> divisors(long n) {
        List<Long> smaller = new ArrayList<>();
        List<Long> larger = new ArrayList<>();

        for (long i = 1; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                smaller.add(i);
                if (n/i != i) {
                    larger.add(n / i);
                }
            }
        }

        Collections.reverse(larger);
        smaller.addAll(larger);
        return smaller;
    }

    public static void main(String [] args) {
        Scanner scanner = new Scanner(System.in);
        long n = scanner.nextLong();
        int k = scanner.nextInt();
        List<Long> divisors = divisors(n);
        if (divisors.size() < k) {
            System.out.println(-1);
        } else {
            System.out.println(divisors.get(k-1));
        }
    }
}