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.*;
  import java.util.*;
  
  public class B {
  	
  	public static void main(String args[]) throws NumberFormatException, IOException {		
  		
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		
  		int t = Integer.parseInt(br.readLine());
  		
  		while(t --> 0) {
  			
  			StringTokenizer st = new StringTokenizer(br.readLine());
  			TreeSet<Integer> breakCount = new TreeSet<Integer>();
  			
  			int a = Integer.parseInt(st.nextToken());
  			int b = Integer.parseInt(st.nextToken());
  			int tot = a + b;
  			
  			//start with alice serving first
  			int awin = (tot/2) + ((tot%2==1) ? 1 : 0);
  			int bwin = tot/2;
  			int breaks = 0;
  			
  			if(awin < a) {
  				
  				breaks += a - awin;
  				
  				for(int i = 0; i <= Math.min(awin, b); i++) {
  					
  					breakCount.add(breaks + 2 * i);
  					
  				}
  				
  			}else if(bwin < b) {
  				
  				breaks += b - bwin;
  				
  				for(int i = 0; i <= Math.min(a, bwin); i++) {
  					
  					breakCount.add(breaks + 2 * i);
  					
  				}
  				
  			}else {
  				
  				for(int i = 0; i <= Math.min(a, b); i++) {
  					
  					breakCount.add(breaks + 2 * i);
  					
  				}
  				
  			}			
  						
  			int tmp = a;
  			a = b;
  			b = tmp;
  			
  			awin = (tot/2) + ((tot%2==1) ? 1 : 0);
  			bwin = tot/2;
  			breaks = 0;
  			
  			if(awin < a) {
  				
  				breaks += a - awin;
  				
  				for(int i = 0; i <= Math.min(awin, b); i++) {
  					
  					breakCount.add(breaks + 2 * i);
  					
  				}
  				
  			}else if(bwin < b) {
  				
  				breaks += b - bwin;
  				
  				for(int i = 0; i <= Math.min(a, bwin); i++) {
  					
  					breakCount.add(breaks + 2 * i);
  					
  				}
  				
  			}else {
  				
  				for(int i = 0; i <= Math.min(a, b); i++) {
  					
  					breakCount.add(breaks + 2 * i);
  					
  				}
  				
  			}
  			
  			System.out.println(breakCount.size());			
  			breakCount.forEach(count -> System.out.print(count + " "));
  			System.out.println();
  			
  		}
  		
  	}
  	
  }
  