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 product, permutations, combinations, accumulate, compress
  #from collections import Counter, defaultdict, deque # deque([iterable[, maxlen]]) #appendleft popleft rotate
  #from heapq import heapify, heappop, heappush # func(heapifiedlist, item)
  #from bisect import bisect_left, bisect_right, insort # func(sortedlist, item)
  # TODO: more coping mechanisms...?
  #from sys import setrecursionlimit
  
  import os, sys
  input = lambda: sys.stdin.buffer.readline().rstrip(b'\r\n')
  
  def dbp(*args, **kwargs): # calling with dbp(locals()) is perfectly cromulent
  	print(*args, file=sys.stderr, **kwargs)
  
  a_str = lambda: input().decode('ascii')
  get_int_list = lambda: [int(x) for x in input().split()]
  join_str = lambda l, ch=' ': ch.join((str(x) for x in l))  # ch is a default, not kwarg
  
  def do_thing():
  	b, a = sorted(get_int_list())
  	n = a+b
  
  	def slow_solve():
  		dbp('case:', a, b)
  		ans = []
  		for p in product((0, 1), repeat=n):
  			score = [0] * 2
  			for i in range(n):
  				turn = i % 2
  				if not p[i]:
  					score[turn]+=1
  				else:
  					score[1-turn]+=1
  			#dbp(p, score)
  			if [a,b] == score or [b,a] == score:
  				dbp('added:', p, score)
  				ans.append(sum(p))
  
  		return join_str(sorted(set(ans)))
  
  	#sr = slow_solve()
  
  	a0 = (n+1)//2
  	agap = a-a0#max(0, a-a0)
  	#dbp("a's:", a, a0, agap)
  
  	b0 = n//2
  	bgap = b-b0#max(0, b-b0)
  	#dbp("b's:", b, b0, bgap)
  
  	mink = min(agap, -bgap)
  	#dbp('min k:', mink)
  	maxk = n-mink+1
  
  	ans = []
  	for k in range(mink, maxk, 2):
  		ans.append(k)
  		if n % 2:
  			if k+1 < maxk:
  				ans.append(k+1)
  		else:
  			if k+2 < maxk:
  				ans.append(k+2)
  	
  	ans = sorted(set(ans))
  
  	return str(len(ans))+'\n' + join_str(ans)
  
  if __name__ == "__main__":
  	maxcc = 1
  	maxcc = int(input().strip())
  	for cc in range(maxcc):
  		print(do_thing())
  