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 Contest1561B
  {
  	static class InputReader {
  		BufferedReader reader;
  		StringTokenizer tokenizer;
  		public InputReader(InputStream stream) {
  			reader = new BufferedReader(new InputStreamReader(stream), 32768);
  			tokenizer = null;
  		}
  		String next() { // reads in the next string
  			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
  				try {
  					tokenizer = new StringTokenizer(reader.readLine());
  				} catch (IOException e) {
  					throw new RuntimeException(e);
  				}
  			}
  			return tokenizer.nextToken();
  		}
  		public int nextInt() { // reads in the next int
  			return Integer.parseInt(next());
  		}
  		public long nextLong() { // reads in the next long
  			return Long.parseLong(next());
  		}
  		public double nextDouble() { // reads in the next double
  			return Double.parseDouble(next());
  			}
  		}
  	static InputReader r = new InputReader(System.in);
  	static PrintWriter pw = new PrintWriter(System.out);
  	public static void main(String[] args)
  	{
  		//LOOK FOR INT OVERFLOW
  		int t = r.nextInt();
  		while (t > 0)
  		{
  			t--;
  			int a = r.nextInt(); int b = r.nextInt();
  			if ((a+b)%2 == 0)
  			{
  				int n = (a+b)/2;
  				int max = Math.min(a, n) + Math.min(b,n);
  				int min = Math.abs(b-a)/2;
  				int k = max - min + 1;
  				pw.println(k/2+1);
  				for (int i = 0; i < k - 1; i +=2)
  				{
  					pw.print(min + i + " ");
  				}
  				pw.println(min + k - 1);
  			}
  			else
  			{
  				int c = Math.max(a, b); int e = Math.min(a, b);  a = c; b = e;
  				int n = (a+b)/2;
  				int max = Math.min(a, n+1) + Math.min(b, n);
  				int min = n - b;
  				int k = max - min  +1;
  				pw.println(k);
  				for (int i= 0; i < k - 1; i ++)
  				{
  					pw.print(min + i + " ");
  				}
  				pw.println(min + k - 1);
  			}
  		}
  		
  		pw.close();
  	}
  }
  