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 Kdivisor
{
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        long n=sc.nextLong();
        long k=sc.nextLong();
        long i;
        ArrayList<Long> fp=new ArrayList<>();
        ArrayList<Long> sp=new ArrayList<>();
        for(i=1;i<=(int)Math.sqrt(n) && k<=n;i++)
        {
            if(n%i==0)
            {
                fp.add(i);
                if(i*i!=n)
                sp.add(n/i);
            }
        }
        if(k-1<fp.size())
        System.out.println(fp.get((int) k-1));
        else if(k>fp.size() && k-fp.size()-1<sp.size())
            System.out.println(sp.get((int) (sp.size()-1-(k-fp.size()-1))));
        else
            System.out.println("-1");
    }
}