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.ArrayList;
import java.util.Collections;

public class P001{
    /*
    * k-th divisor
    * http://codeforces.com/problemset/problem/762/A
    */
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        int k = sc.nextInt();
        ArrayList<Long> answer = new ArrayList<Long>();
        for(long i=1;i*i<=n;i++){
            if (n%i==0){
                answer.add(i);
                if (i*i!=n)
                    answer.add(n/i);
            }
        }
        Collections.sort(answer);
        if (k > answer.size()){
            System.out.println(-1);
        }
        else{
            System.out.println(answer.get(k-1));
        }
	}
}