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.Scanner;

public class Main {
    public static void main(String[] args) {
        ArrayList<Long> dividers = new ArrayList<>();
        int counter =-1;
        Scanner scanner =  new Scanner(System.in);
        long n = scanner.nextLong();
        int k = scanner.nextInt();
        int l = (int) (Math.sqrt(n)+1);

        for(long i = 1;i<l;i++){
            if(i==Math.sqrt(n)){
                dividers.add(dividers.size()/2,i);
            }else if(n%i==0){
                counter++;
                dividers.add(counter,i);
                dividers.add(dividers.size()-counter,n/i);
            }
        }
        dividers.add(n);
        if(k>=dividers.size()){
            System.out.println(-1);
        }else{
            System.out.println(dividers.get(k-1));
        }
    }
}