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) {
  		
  		FastReader f = new FastReader();
  		StringBuffer sb=new StringBuffer();
  		
  		int test=f.nextInt();
  		while(test-->0)
  		{	
  			TreeSet<Integer> set=new TreeSet<>();
  			int a=f.nextInt(),b=f.nextInt();
  			if(a==0 || b==0)
  			{
  				set.add((int)Math.ceil((a+b)/2.0));
  				set.add((int)Math.floor((a+b)/2.0));
  			}
  			else if(Math.abs(a-b)==1)
  			{
  				for(int i=0;i<=a+b;i++)
  					set.add(i);
  			}
  			else 
  			{
  				int min=((a+b)/2);
  				int start=min-Math.min(a, b);
  				for(int i=start,j=Math.min(a, b);j>=0;i+=2,j--)
  					set.add(i);
  				min=((a+b+1)/2);
  				start=min-Math.min(a, b);
  				for(int i=start,j=Math.min(a, b);j>=0;i+=2,j--)
  					set.add(i);
  			}
  			sb.append(set.size()+"\n");
  			for(int i:set)
  				sb.append(i+" ");
  			sb.append("\n");
  		}
  		System.out.println(sb);
  	}
  	
  	static 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(); 
  	    } 
  	    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; 
  	    } 
  	} 
  }
  