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.List;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        long n = in.nextLong();
        int k = in.nextInt();
        in.close();

        long dCount = (long) Math.sqrt(n);
        List<Long> divisors = new ArrayList<>();

        for (int i = 1; i <= dCount; i++) {
            if (n % i == 0) {
                divisors.add((long) i);
            }
        }
        if (k <= divisors.size()) {
            System.out.println(divisors.get(k - 1));
        } else {
            int offset = k - divisors.size();
            int isSquareNum = (n - dCount * dCount != 0 ? 0 : 1);
            if (offset > divisors.size() - isSquareNum) {
                System.out.println(-1);
            } else {
                System.out.println(n / divisors.get(divisors.size() - offset - isSquareNum));
            }
        }
    }
}