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 sys
  read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()
  import bisect,string,math,time,functools,random,fractions
  from bisect import*
  from heapq import heappush,heappop,heapify
  from collections import deque,defaultdict,Counter
  from itertools import permutations,combinations,groupby
  rep=range;R=range
  def I():return int(input())
  def LI():return [int(i) for i in input().split()]
  def LI_():return [int(i)-1 for i in input().split()]
  def S_():return input()
  def IS():return input().split()
  def LS():return [i for i in input().split()]
  def NI(n):return [int(input()) for i in range(n)]
  def NI_(n):return [int(input())-1 for i in range(n)]
  def NLI(n):return [[int(i) for i in input().split()] for i in range(n)]
  def NLI_(n):return [[int(i)-1 for i in input().split()] for i in range(n)]
  def StoLI():return [ord(i)-97 for i in input()]
  def ItoS(n):return chr(n+97)
  def LtoS(ls):return ''.join([chr(i+97) for i in ls])
  def RLI(n=8,a=1,b=10):return [random.randint(a,b)for i in range(n)]
  def RI(a=1,b=10):return random.randint(a,b)
  def INP():
      N=8
      n=random.randint(2,N)
      n=6
      a=[(random.randint(1,9),random.randint(1,9)) for i in range(n)]
      #A=[random.randint(1,n) for i in range(m)]
      return a
  def Rtest(T):
      case,err=0,0
      for i in range(T):
          inp=INP()
          a1=naive(inp)
          a2=solve(inp)
          if a1!=a2:
              print(inp)
              
              print('naive',a1)
              print('solve',a2)
              err+=1
          case+=1
      print('Tested',case,'case with',err,'errors')
  def GI(V,E,ls=None,Directed=False,index=1):
      org_inp=[];g=[[] for i in range(V)]
      FromStdin=True if ls==None else False
      for i in range(E):
          if FromStdin:
              inp=LI()
              org_inp.append(inp)
          else:
              inp=ls[i]
          if len(inp)==2:a,b=inp;c=1
          else:a,b,c=inp
          if index==1:a-=1;b-=1
          aa=(a,c);bb=(b,c);g[a].append(bb)
          if not Directed:g[b].append(aa)
      return g,org_inp
  def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1):
      #h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1) # sample usage
      mp=[boundary]*(w+2);found={}
      for i in R(h):
          s=input()
          for char in search:
              if char in s:
                  found[char]=((i+1)*(w+2)+s.index(char)+1)
                  mp_def[char]=mp_def[replacement_of_found]
          mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
      mp+=[boundary]*(w+2)
      return h+2,w+2,mp,found
  def TI(n):return GI(n,n-1)
  def accum(ls):
      rt=[0]
      for i in ls:rt+=[rt[-1]+i]
      return rt
  def bit_combination(n,base=2):
      rt=[]
      for tb in R(base**n):s=[tb//(base**bt)%base for bt in R(n)];rt+=[s]
      return rt
  def gcd(x,y):
      if y==0:return x
      if x%y==0:return y
      while x%y!=0:x,y=y,x%y
      return y
  def YN(x):print(['NO','YES'][x])
  def Yn(x):print(['No','Yes'][x])
  def show(*inp,end='\n'):
      if show_flg:print(*inp,end=end)
  
  mo=10**9+7
  #mo=998244353
  inf=float('inf')
  FourNb=[(-1,0),(1,0),(0,1),(0,-1)];EightNb=[(-1,0),(1,0),(0,1),(0,-1),(1,1),(-1,-1),(1,-1),(-1,1)];compas=dict(zip('WENS',FourNb));cursol=dict(zip('LRUD',FourNb))
  alp=[chr(ord('a')+i)for i in range(26)]
  #sys.setrecursionlimit(10**7)
  
  
  show_flg=False
  show_flg=True
  
  ans=0
  def solve(a,b):
      n=a+b
      rt=set()
      
      x,y=-~n//2,n//2
      for k in range(n+1):
          s=x-a
          t=y-b
          if s+t!=0:
              continue
          if 0<=k+s<=2*x and 0<=k-s<=2*y and (k+s)%2==0 and (k-s)%2==0:
              rt.add(k)
      
      y,x=-~n//2,n//2
      for k in range(n+1):
          s=x-a
          t=y-b
          if s+t!=0:
              continue
          if 0<=k+s<=2*x and 0<=k-s<=2*y and (k+s)%2==0 and (k-s)%2==0:
              rt.add(k)
      
      return sorted(rt)
      
      
  for _ in range(I()):
      ans=0
      a,b=LI()
      ans=solve(a,b)
      
      print(len(ans))
      print(*ans)
      
  
  
  
  
      
  