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 os
  import sys
  from io import BytesIO, IOBase
  #import sys
  #input = sys.stdin.buffer.readline
  BUFSIZE = 8192
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  #######################################
  from itertools import permutations
  from collections import *
  from collections import deque
  from operator import itemgetter , attrgetter
  from decimal import *
  import bisect
  import math
  import heapq as hq
  #import sympy
  MOD=10**9 +7
  
  
  def is_prime(n):
    if n == 2 or n == 3: return True
    if n < 2 or n%2 == 0: return False
    if n < 9: return True
    if n%3 == 0: return False
    r = int(n**0.5)
    # since all primes > 3 are of the form 6n ± 1
    # start with f=5 (which is prime)
    # and test f, f+2 for being prime
    # then loop by 6.
    f = 5
    while f <= r:
  
      if n % f == 0: return False
      if n % (f+2) == 0: return False
      f += 6
    return True
  
  def pow(a,b,m):
      ans=1
      while b:
          if b&1:
              ans=(ans*a)%m
          b//=2
          a=(a*a)%m
      return ans
  #vis=[]
  #graph=[]
  
  def ispalindrome(s):
      if s[:]==s[::-1]:
          return 1
      return 0
  dp=[]
  limit=[]
  v=[]
  def dpdfs(u,t=-1):
      dp[0][u]=0
      dp[1][u]=0
      for i in v[u]:
          if i==t:
  
              continue
          if dp[1][i]==-1:
              dpdfs(i,u)
          dp[0][u]+=max(abs(limit[0][u]-limit[1][i])+dp[1][i],abs(limit[0][u]-limit[0][i])+dp[0][i])
          dp[1][u] += max(abs(limit[1][u] - limit[1][i]) + dp[1][i], abs(limit[1][u] - limit[0][i]) + dp[0][i])
  vis=[]
  f=0
  def dfs(i):
      vis[i]=1
      act[i]=1
      for j in v[i]:
          if act[j]:
              f=1
              #print(-1)
              return -1
          if vis[j]==0:
             if dfs(j)==-1:
                 return -1
      act[i]=0
      ans.append(i)
      return 0
  from queue import PriorityQueue
  
  def z_algorithm(s):
      res = [0] * len(s)
      res[0] = len(s)
      i, j = 1, 0
      while i < len(s):
          while i + j < len(s) and s[j] == s[i + j]:
              j += 1
          res[i] = j
          if j == 0:
              i += 1
              continue
          k = 1
          while i + k < len(s) and k + res[k] < j:
              res[i + k] = res[k]
              k += 1
          i, j = i + k, j - k
      return res
  def gcd(a, b):
      if a == 0:
          return b
      return gcd(b % a, a)
  
  
  # Function to return LCM of two numbers
  def lcm(a, b):
      return (a / gcd(a, b)) * b
  
  
  def gcdExtended(a, b):
      # Base Case
      if a == 0:
          return b, 0, 1
  
      gcd, x1, y1 = gcdExtended(b % a, a)
  
      # Update x and y using results of recursive
      # call
      x = y1 - (b // a) * x1
      y = x1
  
      return gcd, x, y
  import re
  for _ in range(int(input())):
      a,b=map(int,input().split())
      if b<a:
          a,b=b,a 
      ans=[]
      total=a+b 
      for i in range(a+1):
          ans.append(a-i+(total+1)//2-i)
          if total%2:
              ans.append(a-i+(total//2)-i)
      ans=list(set(ans))
      ans.sort()
      print(len(ans))
      print(*ans)