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: // package codeForces.round740;
  
  import java.io.BufferedReader;
  import java.io.InputStreamReader;
  import java.util.HashSet;
  import java.util.Set;
  import java.util.StringTokenizer;
  import java.util.TreeSet;
  
  public class TestB {
  	public static void main(String args[]) throws Exception {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		int t = Integer.parseInt(br.readLine().trim());
  		var sb = new StringBuilder();
  		while(t-- > 0){
  			int arr[] = nextIntArray(2, br);
  			sb.append(solveQuery(arr)).append("\n");
  		}
  		br.close();
  		System.out.println(sb);
  	}
  
  	private static String solveQuery(int arr[]){
  		var sb = new StringBuilder();
  		Set<Integer> hSet = new TreeSet<>();
  		swap(arr);
  
  		int total = arr[0] + arr[1];
  		int first = (int)Math.floor(((double)total) / 2);
  		int second = total - first;
  		
  
  		// A Goes First
  		for(int i = 0; i <= Math.min(arr[1], first); i++){
  			int otherBreaks = arr[0] - (first - i);
  			hSet.add(i+otherBreaks);
  		}
  
  		// A Decides Breaks
  		for(int i = 0; i <= Math.min(arr[1], second); i++){
  			int otherBreaks = arr[0] - (second - i);
  			hSet.add(i + otherBreaks);
  		}
  
  		sb.append(hSet.size()).append("\n");
  		for(int i: hSet) sb.append(i).append(" ");
  		return sb.toString();
  	}
  
  	private static void swap(int arr[]){
  		if(arr[0] < arr[1]){
  			int temp = arr[0];
  			arr[0] = arr[1];
  			arr[1] = temp;
  		}
  	}
  
  	private static int[] nextIntArray(int N, BufferedReader br) throws Exception {
  		StringTokenizer st = new StringTokenizer(br.readLine().trim()," ");
  		int arr[] = new int[N];
  		for(int i = 0; i < N; i++)
  			arr[i] = Integer.parseInt(st.nextToken());
  		return arr;
  	}
  }
  