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.util.*;
  import java.io.*;
  
  public class B_Charmed_by_the_Game{
      public static void main(String[] args) {
          FastScanner s= new FastScanner();
          StringBuilder res = new StringBuilder();
          int t=s.nextInt();
          int p=0;
          while(p<t){
          int a=s.nextInt();
          int b=s.nextInt();
          int total=a+b;
          ArrayList<Integer> list = new  ArrayList<Integer>();
          HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
          int alice=total/2;
          int bob=total-alice;
          int less=Math.min(a,b);
          int num=0;
          int num2=0;
          int num3=0;
  
          if(less==a){
              num=a;
              num2=b;
              num3=bob;
          }
          else{
              num=b;
              num2=a;
              num3=alice;
          }
          for(int i=less;i>=0;i--){
              int count=0;
              int yo=less-i;
              count+=yo;
              int cc=num2-(num3-yo);
              count+=cc;
              if(!map.containsKey(count)){
                  list.add(count);
                  map.put(count,1);
              }
            //  System.out.println(count);
          }
          int temp=alice;
          alice=bob;
          bob=temp;
           less=Math.min(a,b);
           num=0;
           num2=0;
           num3=0;
  
          if(less==a){
              num=a;
              num2=b;
              num3=bob;
          }
          else{
              num=b;
              num2=a;
              num3=alice;
          }
          for(int i=less;i>=0;i--){
              int count=0;
              int yo=less-i;
              count+=yo;
              int cc=num2-(num3-yo);
              count+=cc;
              if(!map.containsKey(count)){
                  list.add(count);
                //  System.out.println(count);
              }
              
          }
          Collections.sort(list);
          res.append(list.size()+"\n");
          for(int i=0;i<list.size();i++){
              res.append(list.get(i)+" ");
          }
          res.append("\n");
          p++;
      }
      System.out.println(res);
      }
      static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
   
  		public FastScanner(String s) {
  			try {
  				br = new BufferedReader(new FileReader(s));
  			} catch (FileNotFoundException e) {
  				// TODO Auto-generated catch block
  				e.printStackTrace();
  			}
  		}
   
  		public FastScanner() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
   
  		String nextToken() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					// TODO Auto-generated catch block
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
   
  		int nextInt() {
  			return Integer.parseInt(nextToken());
  		}
   
  		long nextLong() {
  			return Long.parseLong(nextToken());
  		}
   
  		double nextDouble() {
  			return Double.parseDouble(nextToken());
  		}
  	}
  
  
  
  
  }