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.*;
public class cuteSeq {

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner ir=new Scanner(System.in);
		
		
		int q = ir.nextInt();
		for (int t = 0; t < q; t++) {
			long a = ir.nextLong();
			long b = ir.nextLong();
			long m = ir.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[] res = new long[(int) k + 1];
			res[0] = a;
			res[(int) k] = b;
			long x = b - (1L << (k - 1)) * (a + 1);
			long tot = a;
			for (int i = 1; i < k; i++) {
				long y = Math.min(m - 1, x / (1L << (long) (k - i - 1)));
				res[i] = tot + y + 1;
				x -= (1L << (long) (k - i - 1)) * y;
				tot += res[i];
			}
			System.out.print(k + 1);
			for (int i = 0; i < k + 1; i++)
				System.out.print(" " + res[i]);
			System.out.println();
		}
	}
		
	}


