Problem: 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.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class CF1166D {
	// 1a
	// 1a + 1c1
	// 2a + 1c1 + 1c2
	// 4a + 2c1 + 1c2 + 1c3
	// 8a + 4c1 + 2c2 + 1c3 + 1c4
	// 16a + 8c1 + 4c2 + 2c3 + 1c4 + 1c5	k=6
	// (2^(k-2))a + (2^(k-3))c1 + (2^(k-4))c2 + (2^(k-5))c3 + (2^(k-6))(c4 + c5)
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int q = Integer.parseInt(br.readLine());
		PrintWriter pw = new PrintWriter(System.out);
		long[] rr = new long[51];
		while (q-- > 0) {
			StringTokenizer st = new StringTokenizer(br.readLine());
			long a = Long.parseLong(st.nextToken());
			long b = Long.parseLong(st.nextToken());
			long m = Long.parseLong(st.nextToken());
			if (a == b) {
				pw.println("1 " + a);
				continue;
			}
			boolean yes = false;
			long p;
			for (int k = 2; (p = 1L << k - 2) <= b / a; k++) {
				long d = b - a * p;
				if (p <= d && d / m <= p && d <= p * m) {
					if (d == p * m)
						for (int i = 1; i < k; i++)
							rr[i] = m;
					else {
						for (int i = 1; i < k; i++)
							rr[i] = 1;
						d -= p;
						for (int i = 1; i < k - 1; i++) {
							p /= 2;
							long r = d / p;
							if (r > m - 1)
								r = m - 1;
							rr[i] += r;
							d -= r * p;
						}
						rr[k - 1] += d;
					}
					pw.print(k + " " + a);
					long sum = a;
					for (int i = 1; i < k; i++) {
						a = sum + rr[i];
						pw.print(" " + a);
						sum += a;
					}
					pw.println();
					yes = true;
					break;
				}
			}
			if (!yes)
				pw.println("-1");
		}
		pw.close();
	}
}