Given a positive integer m, we say that a sequence x_1, x_2, ..., x_n of positive integers is m-cute if for every index i such that 2 ≤ i ≤ n it holds that x_i = x_{i - 1} + x_{i - 2} + ... + x_1 + r_i for some positive integer r_i satisfying 1 ≤ r_i ≤ m.

You will be given q queries consisting of three positive integers a, b and m. For each query you must determine whether or not there exists an m-cute sequence whose first term is a and whose last term is b. If such a sequence exists, you must additionally find an example of it.

Input

The first line contains an integer number q (1 ≤ q ≤ 10^3) — the number of queries.

Each of the following q lines contains three integers a, b, and m (1 ≤ a, b, m ≤ 10^{14}, a ≤ b), describing a single query.

Output

For each query, if no m-cute sequence whose first term is a and whose last term is b exists, print -1.

Otherwise print an integer k (1 ≤ k ≤ 50), followed by k integers x_1, x_2, ..., x_k (1 ≤ x_i ≤ 10^{14}). These integers must satisfy x_1 = a, x_k = b, and that the sequence x_1, x_2, ..., x_k is m-cute.

It can be shown that under the problem constraints, for each query either no m-cute sequence exists, or there exists one with at most 50 terms.

If there are multiple possible sequences, you may print any of them.

Example

Input


2
5 26 2
3 9 1


Output


4 5 6 13 26
-1

Note

Consider the sample. In the first query, the sequence 5, 6, 13, 26 is valid since 6 = 5 + \bf{\color{blue} 1}, 13 = 6 + 5 + {\bf\color{blue} 2} and 26 = 13 + 6 + 5 + {\bf\color{blue} 2} have the bold values all between 1 and 2, so the sequence is 2-cute. Other valid sequences, such as 5, 7, 13, 26 are also accepted.

In the second query, the only possible 1-cute sequence starting at 3 is 3, 4, 8, 16, ..., which does not contain 9.
import java.util.Scanner;
public class D {
    public static void main(String[] args) {
       Scanner sc=new Scanner(System.in);
        int q=sc.nextInt();
        for (int test = 0; test < q; test++) {
            long a=sc.nextLong(),b=sc.nextLong(),m=sc.nextLong();
            if(a==b)System.out.println("1 "+b);
            else{
            long sum=a+1;
            long lentgh;
            for(lentgh=0;sum<=b;lentgh++){
                sum<<=1;
            }
            m--;
            long last=sum>>1;
            long k=a+1;
            long sh=1L<<(lentgh-1);
            if(sh>=(long) Math.ceil((double) (b-last)/m)){
                System.out.print(lentgh+1+" ");
                long d;
                sum=0;
                long msum=0;
                System.out.print(a+" ");
                for(;sh>1;k<<=1,msum+=sum,msum<<=1,sum=d,last+=d*sh) {
                    sh>>=1;
                    d=Math.min((b-last)/sh,m);
                    System.out.print(k+sum+msum+d+" ");
                }
                System.out.println(b);
            }
            else System.out.println(-1);
        }}
    }
}
