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: from itertools import combinations, permutations
  from sys import path_hooks, stdin, stdout
  from collections import Counter,deque
  import math
  from copy import deepcopy
  import random
  import heapq
  import sys
  sys.setrecursionlimit(10**5)
  
   
  def mapinput():
      return  map(int, stdin.readline().split())
  
  def strinput():
      return stdin.readline().strip()
  
  def listinput():
      return list(map(int,stdin.readline().split()))
  
  def intinput():
      return int(stdin.readline().strip())
  
  def gcd(a,b):
      if(b==0):
          return a
      else:
          return gcd(b,a%b)
  
  def SieveOfEratosthenes(n):
      prime = [True for i in range(n + 1)]
      p = 2
      while (p * p <= n):
          if (prime[p] == True):
              for i in range(p * 2, n + 1, p):
                  prime[i] = False
          p += 1
      prime[0]= False
      prime[1]= False
      ans = []
      for p in range(n + 1):
          if prime[p]:
              ans.append(p)
      return ans
  
  coprime1 = {6: 100003, 5: 10007, 4: 1009, 3: 101, 2: 11, 1: 2, 7: 1000003, 8: 10000019, 9: 100000007}
  coprime2 = {8:10000169,9:100000049,7:1000033,6:100019,5:10009,4:1013,3:103,2:13,1:3}
   
  def writ(ss):
      stdout.write(str(ss) + "\n")
  
  mod =  998244353
  
  def sumn(n):
      return (n*(n+1))//2
  
  
  def perm(lis):
      if len(lis) < 2:
          return lis
      #print(lis)
      anss = []
      for i in range(len(lis)):
          ele = lis[i]
          liss = lis[:i] + lis[i+1:]
          for j in perm(liss):
              anss.append( ele+j )
      return anss
  
  for test in range(intinput()):
      def solve():
          #n = intinput()
          a , b = mapinput()
          if a + b & 1 == 0:
              n = (a + b) >> 1
              aa = abs(  (n & 1) - (a & 1)  )
              ans = deque([i for i in range( aa , a + b + 1 , 2 )])
              diff = abs(a - b)>> 2
              for i in range(diff):
                  ans.popleft()
                  ans.pop()
              
  
          else:
              ans = deque([ i for i in range(a+b+1)])
              diff = abs( a - b) >> 1
              for i in range(diff):
                  ans.popleft()
                  ans.pop()
          print(len(ans))
          print(*ans)
  
  
  
  
  
  
  
      solve()
  
  