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 Main {

	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();
			long b = sc.nextLong();
			long m = sc.nextLong();
			
			long[] result = new long[200];
			long sum = a;
			result[0] = a;
			int n = 1;
			// Генерируем последовательность, где все r = 1
			for (n = 1; result[n - 1] <= b; n++) {
				result[n] = sum + 1;
				sum += result[n];
			}
			// n - номер последнего элемента
			n--;
			
			for (int i = 1; i < n - 1; i++) {
				long t = Math.min(m - 1, (b - result[n - 1]) / (long)(Math.pow(2, n - 2 - i)));
				result[i] += t;
				for (int j = i + 1; j < n; j++) {
					result[j] += t * (long)(Math.pow(2, j - 1 - i));
				}
			}
			result[n - 1] += Math.min(m - 1, b - result[n - 1]);
			
			if (result[n - 1] == b) {
				System.out.print(n);
				for (int i = 0; i < n; i++) {
					System.out.print(" " + result[i]);
				}
				System.out.println();
			} else {
				System.out.println(-1);
			}
		}
	}
}
