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

public class Main {
    public static void main(String[] args) throws java.lang.Exception {
        Scanner in=new Scanner(System.in);
        int q = in.nextInt();
        for (int t=0;t<q;t++) {
            long a=in.nextLong();
            long b=in.nextLong();
            long m=in.nextLong();
            long k=0;
            if(a==b) {
                System.out.println("1 "+a);
                continue;
            }
            while((1L<<(k-1))*(a+m)<b)
                k++;
            if((1L<<(k-1))*(a+1)>b) {
                System.out.println(-1);
                continue;
            }
            long[] result = new long[(int)(k+1)];
            result[0]=a;
            result[(int)k] = b;
            long x = b-(1L<<(k-1))*(a+1);
            long total=a;
            for (int i=1;i<k;i++) {
                long y = Math.min(m-1, x/(1L<<(long)(k-i-1)));
                result[i]=total+y+1;
                x-=(1L<<(long)(k-i-1))*y;
                total+=result[i];
            }
            System.out.print(k+1);
            for (int i=0;i<=k;i++)
                System.out.print(" " + result[i]);
            System.out.println();
        }
        }
}
