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 D {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int q = in.nextInt();
        long s[][] = new long[q][3];
        for (int i = 0; i < q; i++) {
            s[i][0] = in.nextLong();
            s[i][1] = in.nextLong();
            s[i][2] = in.nextLong();
        }
        long l, r, a, b, m, k;
        int power;
        for (int i = 0; i < q; i++) {
            if (s[i][0] == s[i][1]) {
                System.out.println(1 + " " +s[i][0]);
            } else {
                k = 1;
                a = s[i][0];
                b = s[i][1];
                m = s[i][2];
                l = a + 1;
                r = a + m;
                power = 0;
                while (k * r < b) {
                    k *= 2;
                    power++;
                }
                if (k * l <= b) {
                    long H = b - k * l;
                    long powers[] = new long[power + 1];
                    k /= 2;
                    for (int j = power; j >= 0; j--) {
                        if (k == 0) {
                            k = 1;
                        }
                        long delta = H / k;
                        long min = Math.min(delta, m - 1);
                        powers[j] = min;
                        H -= min * k;
                        k /= 2;
                    }
                    long z = s[i][0];
                    System.out.print(power + 2 + " ");
                    System.out.print(z + " ");
                    for (int j = power; j >= 0; j--) {
                        System.out.print(z + powers[j] + 1 + " ");
                        z += z + powers[j] + 1;
                    }
                    System.out.println();
                } else {
                    System.out.println(-1);
                }
            }
        }
    }
}
