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.Scanner;
import java.util.TreeSet;

public class KthDivisor {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        TreeSet<Long> set = new TreeSet<>();
        long n = scanner.nextLong();
        long k = scanner.nextInt();
        set.add(1L);
        set.add(n);
        for(long i = 2; i*i <= n; i++)
            if (n % i == 0) {
                set.add(i);
                set.add(n / i);
            }
        if (k > set.size())
            System.out.println(-1);
        else {
            int iter = 1;
            for(long i : set) {
                if (iter == k) System.out.println(i);
                iter++;
            }
        }
    }
}