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 math
  from itertools import permutations
  
  
  t = int(input())
  for i in range(t):
      a, b = [int(x) for x in input().split()]
      posibles = set()
      p = []
      for i in range(a):
          p.append(1)
      for i in range(b):
          p.append(0)
      if abs(a - b) == 1:
          for i in range(a+b+1):
              posibles.add(i)
      elif a == b:
          for i in range(0,a+b+1, 2):
              posibles.add(i)
      elif (a % 2 == 1 and b % 2 == 1)or(a % 2 == 0 and b % 2 == 0):
          n = 0
          tmp = 0
          if (a > b):
              n = (a - b)//2
              tmp = b//2+1
          else:
              n = (b - a)//2
              tmp = a//2+1
          tmp = tmp*2
          counter = 0 + (a % 2 == 0 and b % 2 == 0)
          for i in range(n, a+b, 2):
              posibles.add(i)
              counter += 1
              if counter == tmp:
                  break 
      else:
          n = 0
          small = 0
          tmp = 0
          if (a > b):
              n = (a - b)//2
              tmp = b//2+1
              small = b
          else:
              n = (b - a)//2
              tmp = a//2+1
              small = a
          counter = 0
          for i in range(n, a+b):
              posibles.add(i)
              counter+= 1
              if counter >= (small+1)*2:
                  break
          
      print(len(posibles))
      s = ""
      p = sorted(posibles)
      for i in range(len(p)-1):
          print(p[i], end=" ")
      print(p[len(p)-1])