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{
  	static FastReader sc=null;
  	
  	public static void main(String[] args) {
  		sc=new FastReader();
  		//ABABABAB
  		//4 4 - [0-8]   d=(a-b) , s=d/2 ,e=n-d/2
  		//3 5 - [1-7]
  		//2 6 - [2-6]
  		//1 7 - [3-5]
  		//0 8 - [4-4]
  		
  		//ABABABA [greater same as even
  		//4 3 - [0-6]
  		
  		
  		//BABABAB
  		//4 3 - [1-7]
  		
  		PrintWriter out=new PrintWriter(System.out);
  		int t=sc.nextInt();
  		while(t-->0) {
  			int a=sc.nextInt(),b=sc.nextInt(),n=(a+b),d=Math.abs(a-b);
  			
  			int min=d/2,max=n-d/2,add=2-n%2;
  			ArrayList<Integer> ans=new ArrayList<>();
  			for(int i=min;i<=max;i+=add)ans.add(i);
  			
  			out.println(ans.size());
  			for(int e:ans)out.print(e+" ");
  			out.println();
  		}
  		out.close();
  	}
  	
  	static int[] ruffleSort(int a[]) {
  		ArrayList<Integer> al=new ArrayList<>();
  		for(int i:a)al.add(i);
  		Collections.sort(al);
  		for(int i=0;i<a.length;i++)a[i]=al.get(i);
  		return a;
  	}
  	
  	static void print(int a[]) {
  		for(int e:a) {
  			System.out.print(e+" ");
  		}
  		System.out.println();
  	}
  	
  	static class FastReader{
  		
  		StringTokenizer st=new StringTokenizer("");
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		
  		String next() {
  			while(!st.hasMoreTokens()) 
  				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());
  		}
  		
  		int[] readArray(int n) {
  			int a[]=new int[n];
  			for(int i=0;i<n;i++)a[i]=sc.nextInt();
  			return a;
  		}
  	}
  	
  	
  }
  