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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class _762A {
    public static void main(String args[]){
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        long n = in.nextLong();
        int k = in.nextInt();
        long sqrt = (long)Math.sqrt(n) + 1;
        ArrayList<Long> divisors = new ArrayList<>();

        for(long i = 1; i < sqrt; i++)
            if(n % i == 0) {
                divisors.add(i);
                if(!divisors.contains(n / i))divisors.add(n / i);
            }
        Collections.sort(divisors);
        if(divisors.size() >= k)
            out.print(divisors.get(--k));
        else out.print(-1);


            out.close();
    }
}