description: Alice and Borys are playing tennis.
  
  A tennis match consists of games. In each game, one of the players is serving and the other one is receiving.
  
  Players serve in turns: after a game where Alice is serving follows a game where Borys is serving, and vice versa.
  
  Each game ends with a victory of one of the players. If a game is won by the serving player, it's said that this player holds serve. If a game is won by the receiving player, it's said that this player breaks serve.
  
  It is known that Alice won a games and Borys won b games during the match. It is unknown who served first and who won which games.
  
  Find all values of k such that exactly k breaks could happen during the match between Alice and Borys in total.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
  
  Each of the next t lines describes one test case and contains two integers a and b (0 ≤ a, b ≤ 10^5; a + b > 0) — the number of games won by Alice and Borys, respectively.
  
  It is guaranteed that the sum of a + b over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case print two lines.
  
  In the first line, print a single integer m (1 ≤ m ≤ a + b + 1) — the number of values of k such that exactly k breaks could happen during the match.
  
  In the second line, print m distinct integers k_1, k_2, …, k_m (0 ≤ k_1 < k_2 < … < k_m ≤ a + b) — the sought values of k in increasing order.
  
  Example
  
  Input
  
  
  3
  2 1
  1 1
  0 5
  
  
  Output
  
  
  4
  0 1 2 3
  2
  0 2
  2
  2 3
  
  Note
  
  In the first test case, any number of breaks between 0 and 3 could happen during the match: 
  
    * Alice holds serve, Borys holds serve, Alice holds serve: 0 breaks; 
    * Borys holds serve, Alice holds serve, Alice breaks serve: 1 break; 
    * Borys breaks serve, Alice breaks serve, Alice holds serve: 2 breaks; 
    * Alice breaks serve, Borys breaks serve, Alice breaks serve: 3 breaks. 
  
  
  
  In the second test case, the players could either both hold serves (0 breaks) or both break serves (2 breaks).
  
  In the third test case, either 2 or 3 breaks could happen: 
  
    * Borys holds serve, Borys breaks serve, Borys holds serve, Borys breaks serve, Borys holds serve: 2 breaks; 
    * Borys breaks serve, Borys holds serve, Borys breaks serve, Borys holds serve, Borys breaks serve: 3 breaks. 
solution: import java.io.PrintWriter;
  import java.util.ArrayList;
  import java.util.Collections;
  import java.util.Scanner;
  
  public class Main {
      
      public static void main(String[] args) {
          
          var sc = new Scanner(System.in);
          var pw = new PrintWriter(System.out);
          
          int T = Integer.parseInt(sc.next());
          for(int t = 0; t < T; t++){
              int a = Integer.parseInt(sc.next());
              int b = Integer.parseInt(sc.next());
              
              if(a > b){
                  int tmp = a;
                  a = b;
                  b = tmp;
              }
              
              var list = new ArrayList<Integer>();
              if((a+b) % 2 == 0){
                  int holdA = a;
                  int breakA = 0;
                  int holdB = (a+b)/2;
                  int breakB = (a+b)/2 - a;
                  list.add(breakB);
                  while(holdA > 0){
                      holdA--;
                      breakA++;
                      holdB--;
                      breakB++;
                      list.add(breakA + breakB);
                  }
              }else{
                  int holdA = a;
                  int breakA = 0;
                  int holdB = (a+b)/2;
                  int breakB = (a+b)/2 + 1 - a;
                  list.add(breakB);
                  while(holdA > 0){
                      holdA--;
                      breakA++;
                      holdB--;
                      breakB++;
                      list.add(breakA + breakB);
                  }
      
                  holdA = a;
                  breakA = 0;
                  holdB = (a+b)/2 + 1;
                  breakB = (a+b)/2 - a;
                  list.add(breakB);
                  while(holdA > 0){
                      holdA--;
                      breakA++;
                      holdB--;
                      breakB++;
                      list.add(breakA + breakB);
                  }
              }
              Collections.sort(list);
              pw.println(list.size());
              for(Integer value : list){
                  pw.print(value + " ");
              }
              pw.println();
          }
          pw.flush();
      }
  }