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.IOException;
import java.io.PrintWriter;
import java.util.Scanner;

// https://codeforces.com/contest/1166/problem/D
public class CuteSequences {
public static void main(String[] xx) throws IOException {
    Scanner sc = new Scanner(System.in);
    PrintWriter pw = new PrintWriter(System.out);
    int q = sc.nextInt();
    long[] twoPow = new long[51];
    twoPow[0] = 1;
    for (int i = 1; i < twoPow.length; i++) {
        twoPow[i] = twoPow[i - 1] * 2;
    }
    for (int i = 0; i < q; i++) {
        long a = sc.nextLong();
        long b = sc.nextLong();
        long m = sc.nextLong();
        boolean found = false;
        for (int k = 2; !found && k <= 4 + Long.numberOfLeadingZeros(a) - Long.numberOfLeadingZeros(b); k++) {
            if (b < twoPow[k - 2] * (a + 1) || b > twoPow[k - 2] * (a + m)) {
                continue;
            }
            found = true;
            long[] rem = new long[k - 1];
            b -= twoPow[k - 2] * a;
            for (int j = 0; j < k - 1; j++) {
                b -= twoPow[Math.max(0, j - 1)];
                rem[j]++;
            }
            for (int j = 0; j < k - 1; j++) {
                rem[j] += b / twoPow[Math.max(0, k - j - 3)];
                rem[j] = Math.min(rem[j], m);
                b -= twoPow[Math.max(k - j - 3, 0)] * (rem[j] - 1);
            }
            pw.print(k + " ");
            pw.print(a);
            long S = a;
            for (int j = 0; j < k - 1; j++) {
                pw.print(" " + (S + rem[j]));
                S = S + S + rem[j];
            }
            pw.println();
        }
        if (a == b) {
            pw.println(1 + " " + a);
        } else if (!found) {
            pw.println(-1);
        }
    }
    pw.flush();
    pw.close();
}
}