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: T = int(input())
  
  def solve():
      a, b = map(int, input().split())
      n = a+b
      if n%2==0:
          c = min(a, b)+1
          print(c)
          print(*[x for x in range(n//2-min(a, b), n//2-min(a, b)+2*c, 2)])
      else:
          start = n//2-min(a, b)
          c = (min(a,b)+1)*2
          print(c)
          print(*[x for x in range(start, start+c)])
  ##    mask1, mask2 = 0, 0
  ##    n = a+b
  ##    for i in range(n):
  ##        if i%2 != 0:
  ##            mask1 |= (1 << i)
  ##        else:
  ##            mask2 |= (1 << i)
  ##    h1, h2 = bin(mask1).count("1"), bin(mask2).count("1")
  ##    set_n = set()
  ##    for m in range((1 << n)):
  ##        if bin(m).count("1") == a or bin(m).count("1") == b:
  ##            k1, k2 = n-bin(mask1^m).count("1"), n-bin(mask2^m).count("1") 
  ##            set_n.add(k1)
  ##            set_n.add(k2)
  ##    print(len(set_n))
  ##    print(' '.join([str(x) for x in sorted(set_n)]))
  ##    1
  ##    2 3
  ##    6
  ##    0 1 2 3 4 5
  ##    >>> 
  ##    >>> 
  ##    ========= RESTART: /Users/apple/Documents/Codeforces/Round 740/B.py =========
  ##    1
  ##    1 6
  ##    4
  ##    2 3 4 5
  ##    >>> 
  ##    ========= RESTART: /Users/apple/Documents/Codeforces/Round 740/B.py =========
  ##    1
  ##    2 5
  ##    6
  ##    1 2 3 4 5 6
  ##    >>> 
  ##    ========= RESTART: /Users/apple/Documents/Codeforces/Round 740/B.py =========
  ##    1
  ##    3 4
  ##    8
  ##    0 1 2 3 4 5 6 7
      
  for _ in range(T):
      solve()
  
  
  
      
  