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: '''
      Auther: ghoshashis545 Ashis Ghosh
      College: Jalpaiguri Govt Enggineering College
  
  '''
  from os import path
  from io import BytesIO, IOBase
  import sys
  from heapq import heappush,heappop
  from functools import cmp_to_key as ctk
  from collections import deque,Counter,defaultdict as dd 
  from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
  from itertools import permutations
  from datetime import datetime
  from math import ceil,sqrt,log,gcd
  def ii():return int(input())
  def si():return input().rstrip()
  def mi():return map(int,input().split())
  def li():return list(mi())
  abc='abcdefghijklmnopqrstuvwxyz'
  mod=1000000007
  #mod=998244353
  inf = float("inf")
  vow=['a','e','i','o','u']
  dx,dy=[-1,1,0,0],[0,0,1,-1]
  
  def bo(i):
      return ord(i)-ord('0')
  
  file = 1
  def ceil(a,b):
      return (a+b-1)//b
  
  
  
  
  def solve():
  
      
      for _ in range(1,ii()+1):
          a,b = mi()
          d = abs(a-b)//2
          if (a+b)&1:
              print((a+b-2*d)+1)
              for i in range(d,a+b-d+1):
                  print(i,end=" ")
          else:
              print((a+b-2*d)//2+1)
              for i in range(d,a+b-d+1,2):
                  print(i,end=" ")
          print()
  
          
  
  
  
  
  
  
  
              
  
  
  
  
  
      
          
  
  
  
  
  
  
  
  
          
  if __name__ =="__main__":
  
      if(file):
      
         if path.exists('input.txt'):
             sys.stdin=open('input.txt', 'r')
             sys.stdout=open('output.txt','w')
         else:
             input=sys.stdin.readline
      solve()
  