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 in = new FastReader();
  		StringBuilder ans = new StringBuilder();
  
  		int T = in.nextInt();
  		while(T-- > 0) {
  			int a = in.nextInt();
  			int b = in.nextInt();
  			
  			TreeSet<Integer> breaks = new TreeSet<>();
  			
  			int p = (int) Math.ceil((a+b)/2.0);
  			int q = a+b-p;
  			for(int x = 0; x <= p; x++) {
  				int y = a - (p-x);
  				if(0<=y && y<=q) breaks.add(x+y);
  			}
  			for(int x = 0; x <= q; x++) {
  				int y = a - (q-x);
  				if(0<=y && y<=p) breaks.add(x+y);
  			}
  
  			ans.append(breaks.size() + "\n");
  			for(int k : breaks) ans.append(k + " ");
  			ans.append("\n");
  		}
  
  		System.out.print(ans);
  	}
  
  	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;
  		}
  		int[] readInt(int size) {
  			int[] arr = new int[size];
  			for(int i = 0; i < size; i++)
  				arr[i] = Integer.parseInt(next());
  			return arr;
  		}
  		long[] readLong(int size) {
  			long[] arr = new long[size];
  			for(int i = 0; i < size; i++)
  				arr[i] = Long.parseLong(next());
  			return arr;
  		}
  		int[][] read2dArray(int rows, int cols) {
  			int[][] arr = new int[rows][cols];
  			for(int i = 0; i < rows; i++) {
  				for(int j = 0; j < cols; j++)
  					arr[i][j] = Integer.parseInt(next());
  			}
  			return arr;
  		}
  	}
  }