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.
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
What do you think? What do you think?
1st on Billboard, what do you think of it
Next is a Grammy, what do you think of it
However you think, I’m sorry, but shit, I have no fcking interest
*******************************
I'm standing on top of my Monopoly board
That means I'm on top of my game and it don't stop
til my hip don't hop anymore
https://www.a2oj.com/Ladder16.html
*******************************
300IQ as writer = Sad!
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class x1166D
   {
      public static void main(String hi[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int T = Integer.parseInt(st.nextToken());
         StringBuilder sb = new StringBuilder();
         outer:while(T-->0)
         {
            st = new StringTokenizer(infile.readLine());
            long A = Long.parseLong(st.nextToken());
            long B = Long.parseLong(st.nextToken());
            long M = Long.parseLong(st.nextToken());
            if(A == B)
               sb.append("1 "+A+"\n");
            else
            {
               for(int len=2; len <= 50; len++)
               {
                  long[] res = new long[len];
                  long sum = A;  res[0] = A;
                  for(int i=1; i < len; i++)
                  {
                     res[i] = sum+1;
                     sum += res[i];
                  }
                  if(res[len-1] > B)
                     continue;
                  //increase to fit
                  long needed = B-res[len-1];
                  for(int dex=1; dex < len; dex++)
                  {
                     long inc = 1L;
                     if(dex+2 < len)
                        inc = 1L << (len-dex-2);
                     long low = 0L; long high = M-1;
                     while(low != high)
                     {
                        long mid = (low+high+1)/2;
                        if(inc <= 1.0*needed/mid)
                           low = mid;
                        else
                           high = mid-1;
                     }
                     res[dex]+=low; long val = 1L;
                     for(int i=dex+1; i < len; i++)
                     {
                        res[i] += low*val;
                        val <<= 1;
                     }
                     needed -= low*inc;
                     if(needed == 0)
                        break;
                  }
                  if(needed == 0)
                  {
                     long max = res[0];
                     for(long x: res)
                        max = Math.max(max, x);
                     if(max <= 100000000000000L)
                     {
                        sb.append(len+" ");
                        for(long x: res)
                           sb.append(x+" ");
                        sb.append("\n");
                        continue outer;
                     }
                  }
               }
               sb.append("-1\n");
            }
         }
         System.out.print(sb);
      }
   }