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.Arrays;
import java.util.Collections;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;
import java.util.Stack;

public class Test {

    public static void main(String[] args) {
        
        Scanner scanner = new Scanner(System.in);
        long n = scanner.nextLong(), k = scanner.nextInt();
        List<Long> div = new ArrayList<>();
        for (long i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                div.add(i);
                if (i*i != n) {
                    div.add(n/i);
                }
            }
        }
        Collections.sort(div);
        if (div.size() < k) {
            System.out.println(-1);
        }else{
            System.out.println(div.get((int)k-1));
        }
    }
}