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.*;
  
  public class A {
  	
  	public static void main(String[] args) {
  		Scanner scan=new Scanner(System.in);
  		int T=scan.nextInt();
  		
  		
  		while (T-->0) {
  			int a=scan.nextInt(),b=scan.nextInt();
  			HashSet<Integer> set=helper(a, a+b);
  			set.addAll(helper(b, a+b));
  			System.out.println(set.size());
  			ArrayList<Integer> ls=new ArrayList<Integer>(set);
  			Collections.sort(ls);
  			for(int i:ls) {
  				System.out.print(i+" ");
  			}System.out.println();
  			
  		}
  		scan.close();
  	}
  	private static HashSet<Integer> helper(int win,int total) {
  		HashSet<Integer> arr=new HashSet<Integer>();
  		int ehalf=total/2;
  		int ohalf=(int)Math.ceil(total/2.0);
  		int lose=total-win;
  		
  //		System.out.println("win="+win+" lo="+lose);////
  //		System.out.println("eh="+ehalf+" oh="+ohalf);/////
  		
  		
  		for(int i=Math.max(0, win-ohalf);i<=Math.min(win, ehalf);i++) {
  //			System.out.println("----i~"+i+" l~"+(lose-(ehalf-i)));/////
  			arr.add(i+(lose-(ehalf-i)));
  		}
  		return arr;
  		
  	}
  }
  
  
  
  
  
  
  
  
  