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.text.DecimalFormat;
  import java.util.Scanner;
  import java.util.*;
  public class Test1 {
  
  
      public static void main(String[] args) {
          Scanner s = new Scanner(System.in);
  //            测试次数
          int testnum = s.nextInt();
  
  
          for (int k = 0; k <testnum; k++) {
  //
              int a=s.nextInt();
              int b=s.nextInt();
  
              int A=Math.max(a,b);
              int B=Math.min(a,b);
              Set<Integer> ans = new TreeSet<>();
  
              int sum=A+B;
              if(A-B<=1){
              for (int i = 0; i <=sum; i++) {
                  if(i%2==0){
                      ans.add(i);
                      ans.add(sum-i);
                  }
              }
                  System.out.println(ans.size());
  
              for(int i:ans){
  
                  System.out.print(i);
                  System.out.print(" ");
              }
                  System.out.println();
              }else{
                  int index=A-(sum+1)/2;
                  for (int i = A-(sum+1)/2; index<=sum/2; index++) {
  
                          ans.add(i);
                          ans.add(sum-i);
                          i+=2;
  
                  }
                  System.out.println(ans.size());
  
                  for(int i:ans){
  
                      System.out.print(i);
                      System.out.print(" ");
                  }
                  System.out.println();
  
              }
  
  
  
  
          }
  
  
      }
  
  
  
  
  
      private static boolean check(int breaks,int a,int b) {
          int sum=a+b;
          int A=(sum+1)/2;
          int B=sum-A;
          if(breaks==a-A){
              return true;
          }
          return false;
  
      }
  }
  