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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  
  public class new1
  {	
  	public static boolean fun(int[] arr, int n) {
  		for(int i = 1; i < n; i++) {
  			if(arr[i] > arr[i + 1]) return false;
  		}
  		return true;
  	}
  	
  		
  	public static void main (String[] args)
  	{
  
  		Scanner s  = new Scanner(System.in);
  		//System.out.println("y892");
  		int t = s.nextInt();
  		for(int i = 0; i < t; i++) {
  			int a = s.nextInt();
  			int b = s.nextInt();
  			int c1 = 0; int c2 = 0;
  			int c3 = 0; int c4 = 0;
  			if((a + b) % 2 == 0) {
  				int val = (a + b) / 2;
  				c1 = val; c2 = val; c3 = val; c4 = val;	
  			}
  			else {
  				int val1 = (a + b) / 2; int val2 = (a + b) - val1;
  				c1 = val1; c2 = val2;
  				c3 = val2; c4 = val1;
  			}
  			ArrayList<Integer> a1 = new ArrayList<Integer>();
  			for(int j = 0; j <= (a + b); j++) {
  				int aa = a - c2 + j;
  				int bb = c1 - b + j;
  				int aaa = a - c4 + j;
  				int bbb = c3 - b + j;
  				if(aa == bb && (aa % 2 == 0)) {
  					int x = aa / 2; int y = j - x;
  					if( x >= 0 && y >= 0 && a >= x && b >= y && c1 >= x && c2 >= y)
  						a1.add(j);
  					continue;
  				}
  				if(aaa == bbb && (aaa % 2 == 0)) {
  					
  					int x = aaa / 2; int y = j - x;
  					if( x >= 0 && y >= 0 && a >= x && b >= y && c3 >= x && c4 >= y)
  						a1.add(j);
  					//a1.add(j);
  					continue;
  				}
  				
  			}
  			Collections.sort(a1);
  			System.out.println(a1.size());
  			for(int j = 0; j < a1.size(); j++) {
  				System.out.print(a1.get(j) + " ");
  			}
  			System.out.println();
  			
  			
  			
  			
  			//System.out.println(it);
  			
  			
  			
  			
  			
  		}
  		
  		
  	}
  
  }
  
  
  
  
  class FastReader {
     BufferedReader br;
     StringTokenizer st;
     public FastReader() {
         br = new BufferedReader(new InputStreamReader(System.in));
     }
     String next() {
         while (st == null || !st.hasMoreElements()) {
             try {
                 st = new StringTokenizer(br.readLine());
             } catch (IOException e) {
                 e.printStackTrace();
             }
         }
         return st.nextToken();
     }
  
     public int nextInt() {
         return Integer.parseInt(next());
     }
  
     long nextLong() {
         return Long.parseLong(next());
     }
  
     double nextDouble() {
         return Double.parseDouble(next());
     }
  
     String nextLine() {
         String str = "";
         try {
             str = br.readLine();
         } catch (IOException e) {
             e.printStackTrace();
         }
         return str;
  }}