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.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        long q = s.nextLong();
        for (int i=0; i<q; i++) {
            long a = s.nextLong(), b = s.nextLong(), m = s.nextLong();

            if (a == b){
                System.out.println(1+" "+a);
                continue;
            }

            long pow = 1;
            int j=1;
            while (pow*a < b){
                if (pow*a + pow <= b && b <= pow*a + pow*m) {
                    System.out.print(j+1 +" "+a+" ");
                    long rem = b - pow*a - pow, sum=a;
                    long[] num = new long[j];

                    for (int k =0; k<j; k++) {
                        if (pow > 1)
                            pow /= 2;
                        num[k] = Math.min(rem / pow + 1, m);
                        rem -= (num[k]-1) * pow;
                        System.out.print(sum + num[k] +" ");
                        sum += sum + num[k];
                    }
                    break;

                }
                pow*=2;
                j++;
            }
            if (pow*a > b)
                System.out.println(-1);
        }
    }
}