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
  def subsets_gen(num):
      def generator_h(no, s):
          if (len(s) == no):
              yield s
          else:
              yield from generator_h(no, s+"0")
              yield from generator_h(no, s+"1")
      yield from generator_h(num, "")
  
  def perms_gen(pole):
      used = [False for i in range(len(pole))]
      def generator_h(l, m):
          if len(m) == len(l):
              yield m
          else:
              for i in range(len(l)):
                  if used[i]:
                      continue
                  else:
                      m.append(l[i])
                      used[i] = True
                      yield from generator_h(l, m)
                      m.pop(); used[i] = False
      yield from generator_h(pole, [])
  
  def fact(n):
      factors = []
      if (n%2 == 0):
          while(n%2 == 0):
              factors.append(2)
              n//=2
      poc = 3
      while(poc*poc<=n):
          while(n%poc == 0):
              factors.append(poc)
              n//=poc
          poc += 2
      if n>1:
          factors.append(n)
      return factors
  
  def gcd (a, b):
      if (a%b == 0): 
          return b 
      else: 
          return gcd(b, a%b)
   
  def lcm (a, b):
      return (a*b//(gcd(a,b)))
  
  def NcK(n, k):
      if (k == 0 or k == n):
          return 1
      return NcK(n-1, k-1)+NcK(n-1, k)
   
  testcases = 1
  testcases = int(input())
  for _ in range(testcases):
      a, b = map(int, input().split())
      if ((a+b)%2 == 1):
          x = max(a, b)-(a+b+1)//2
          y = a+b - (max(a, b)-(a+b+1)//2)
          print(y-x+1)
          for i in range(x, y+1):
              if (i == y):
                  print(i)
              else:
                  print(i, end = " ")
      else:
          x = max(a, b)-(a+b)//2
          y = a+b - (max(a, b)-(a+b)//2)
          print(math.ceil((y-x+1)/2))
          for i in range(x, y+1, 2):
              if (i == y):
                  print(i)
              else:
                  print(i, end = " ")
  