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.
//JDope
import java.util.*;
import java.io.*;
import java.math.*;
public class D{
public static void main(String[] omkar) throws Exception
{
   BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
   StringTokenizer st = new StringTokenizer(in.readLine());
   StringBuilder sb = new StringBuilder();
   int cases = Integer.parseInt(st.nextToken());
   for(int i = 0; i < cases; i++)
   {
      solve(in, st, sb);
   }  
   System.out.println(sb);
        }
 public static void solve(BufferedReader in, StringTokenizer st, StringBuilder sb) throws Exception
 {
   st = new StringTokenizer(in.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\n");
      return;
   }
   if(a == b)
   {
      sb.append("1 ");
      sb.append(a + "\n");
      return;  
   }
   double fl = (double)(b)/(double)(a+1);
   double ce = (double)(b)/(double)(a+m);
   int n1 = 0;
   double cc = (double)(b);
   while(cc > a+m)
   {
      n1++;
      cc = cc/2.0;
   }
   if(cc < a+1)
   {
      sb.append("-1\n");
      return;
   }
   long[] r = new long[n1+2];
   r[0] = a;
   b = b - (long)(Math.pow(2, n1))*(long)(a+1);
   for(int i = n1-1; i >= 0; i--)
   {
      r[n1-i] = (long)(1 + Math.min((long)((long)(b)/(long)(Math.pow(2, i))), m-1));
      b -= (long)(r[n1-i]-1)*(long)(Math.pow(2, i));
   }
   r[n1+1] = (long)(Math.min(1+b, m));
   b -= (r[n1+1]-1);
   long[] arr = new long[n1+2];
   arr[0] = a;
   for(int i = 1; i < n1+2; i++)
   {
      arr[i] = (long)(2)*(long)(arr[i-1])-(long)(r[i-1])+(long)(r[i]);
   }
   sb.append((n1+2) + " ");
   for(int i = 0; i < n1+2; i++)
   {
      sb.append(arr[i] + " ");
   }
   sb.append("\n");
 }


}