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 MainClass {
      public static void main(String[] args) {
          Reader in = new Reader(System.in);
          StringBuilder stringBuilder = new StringBuilder();
          int t = in.nextInt();
          while (t-- > 0) {
              long a = in.nextLong(), b = in.nextLong();
              long n = a + b;
              TreeSet<Long> ans = new TreeSet<>();
              if (n % 2 == 0) {
                  long max = Math.max(a, b);
                  long minBreaks = max - n / 2;
                  long maxBreaks = minBreaks + 2 * (n / 2 - minBreaks);
                  for (long i = minBreaks; i <= maxBreaks; i += 2) {
                      ans.add(i);
                  }
              } else {
                  long x = n / 2;
                  long y = (n + 1) / 2;
                  solve(a, b, x, y, ans);
                  solve(a, b, y, x, ans);
              }
              stringBuilder.append(ans.size()).append("\n");
              for (long i : ans) {
                  stringBuilder.append(i).append(" ");
              }
              stringBuilder.append("\n");
          }
          System.out.println(stringBuilder);
      }
      public static void solve(long a, long b, long ones, long zeros, TreeSet<Long> ans) {
          long n = a + b;
          long maxBreaks = 0L;
          long minBreaks = 0L;
          if (b > zeros) {
              minBreaks = b - zeros;
              maxBreaks = minBreaks + 2L * a;
          } else {
              minBreaks = a - ones;
              maxBreaks = minBreaks + 2L * Math.min(ones, b);
          }
          for (long i = minBreaks; i <= maxBreaks; i += 2) {
              ans.add(i);
          }
      }
  }
  class Reader {
      public BufferedReader reader;
      public StringTokenizer tokenizer;
  
      public Reader(InputStream stream) {
          reader = new BufferedReader(new InputStreamReader(stream), 32768);
          tokenizer = null;
      }
  
      public String next() {
          while (tokenizer == null || !tokenizer.hasMoreTokens()) {
              try {
                  tokenizer = new StringTokenizer(reader.readLine());
              } catch (IOException e) {
                  throw new RuntimeException(e);
              }
          }
          return tokenizer.nextToken();
      }
  
      public int nextInt() {
          return Integer.parseInt(next());
      }
  
      public long nextLong() {
          return Long.parseLong(next());
      }
  
  }