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 Round740Div2;
  
  import java.io.*;
  import java.util.StringTokenizer;
  
  public class CharmedByTheGame {
      //Upsolving:-
      // we will notice a pattern
      // we need to get total number of games played and the number of serves for each player
      // there is a chance that one player will play that the other player with one game and that if the total number of games was odd
  
      public static void main(String[] args) throws IOException {
          Scanner sc= new Scanner(System.in);
          PrintWriter pw = new PrintWriter(System.out);
          int t = sc.nextInt();
          while(t-->0){
              int a = sc.nextInt();
              int b = sc.nextInt();
              int d = Math.abs(a-b); //difference
              int n = a+b; //total number of games played
              // we will assume that A-->p , B-->q number of games
              int p =(int) Math.ceil((n*1.0)/(2*1.0));
              int q = n/2;
              d/=2;
              int count=0;
              StringBuilder br = new StringBuilder();
              if(n%2==0){ //number of games are even ; A&B played the same number of games
                  for(int i=0;i+d<=n-d;i+=2){
                           br.append((d+i)+" ");
                           count++;
                  }
              }else{
                  for(int i=0;i+d<=n-d;i++){
                           br.append((d+i)+" ");
                           count++;
                  }
              }
              pw.println(count);
              pw.println(br.toString());
  }
          pw.flush();
          pw.close();
      }
      static class Scanner {
          StringTokenizer st;
          BufferedReader br;
  
          public Scanner(InputStream s) {
              br = new BufferedReader(new InputStreamReader(s));
          }
  
          public Scanner(FileReader r) {
              br = new BufferedReader(r);
          }
  
          public String next() throws IOException {
              while (st == null || !st.hasMoreTokens())
                  st = new StringTokenizer(br.readLine());
              return st.nextToken();
          }
  
          public int nextInt() throws IOException {
              return Integer.parseInt(next());
          }
  
          public long nextLong() throws IOException {
              return Long.parseLong(next());
          }
  
          public String nextLine() throws IOException {
              return br.readLine();
          }
  
          public double nextDouble() throws IOException {
              String x = next();
              StringBuilder sb = new StringBuilder("0");
              double res = 0, f = 1;
              boolean dec = false, neg = false;
              int start = 0;
              if (x.charAt(0) == '-') {
                  neg = true;
                  start++;
              }
              for (int i = start; i < x.length(); i++)
                  if (x.charAt(i) == '.') {
                      res = Long.parseLong(sb.toString());
                      sb = new StringBuilder("0");
                      dec = true;
                  } else {
                      sb.append(x.charAt(i));
                      if (dec)
                          f *= 10;
                  }
              res += Long.parseLong(sb.toString()) / f;
              return res * (neg ? -1 : 1);
          }
          public long[] nextlongArray(int n) throws IOException {
              long[] a = new long[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextLong();
              return a;
          }
  
          public Long[] nextLongArray(int n) throws IOException {
              Long[] a = new Long[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextLong();
              return a;
          }
  
          public int[] nextIntArray(int n) throws IOException {
              int[] a = new int[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextInt();
              return a;
          }
  
          public Integer[] nextIntegerArray(int n) throws IOException {
              Integer[] a = new Integer[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextInt();
              return a;
          }
  
  
          public boolean ready() throws IOException {
              return br.ready();
          }
  
      }
  }
  