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.io.*;
import java.util.*;
public class D {
    public static void main(String args[]) throws Exception{
        Scanner in = new Scanner(System.in);
        int q = in.nextInt();
        for (int i = 0; i < q; i++) {
            long a = in.nextLong();
            long b = in.nextLong();
            long m = in.nextLong();
            if (a == b) {
                System.out.println("1 " + a);
                continue;
            }

            long ratio = b/(a+1);
            long power = (int)(Math.log(ratio)/Math.log(2));
            long last = ((1+a) * (long) Math.pow(2, power));
            long diff = b - last;
//            if (diff < Math.pow(2, power)) {
//
//            }
            if (diff > Math.pow(2, power) * (m-1)) {
                System.out.println(-1);
                continue;
            }
            int seqSize = (int) power + 2;
            System.out.print(seqSize + " " + a);
            long[] ms = new long[seqSize + 1];
            for (int j = seqSize; j >= 2; j--) {
                long newM = Math.min((j==2) ? diff : (long) (diff / Math.pow(2, j - 3)), m - 1);
                ms[seqSize - j + 2] = newM+1;
                diff -= newM * Math.pow(2, j - 3);
            }
//            System.out.println(Arrays.toString(ms));
            long prevM = 0;
            for (int j = 2; j <= seqSize; j++) {
                System.out.print(" " + ((long)Math.pow(2, j-2) * a + prevM + ms[j]));
                prevM = 2 * prevM + ms[j];
            }
            System.out.println();
        }
    }
}
