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. 
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())
    ans=[]
    total=a+b
    if min(a,b)==total//2 and max(a,b)==((total+1)//2):
        ans.append(0)
    j=1
    mino=total//2
    maxo=(total+1)//2
    while j<=total:
        if total%2==0 and j%2==1:
            j+=1
            continue
        if total%2==1 and j%2==1:
            if a>b:
                a,b=b,a
            x=min(mino,a)+min(maxo,b)
            y=max(abs(mino-a),abs(maxo-b)) 
            if min(x,y)<=j and max(x,y)>=j:
                ans.append(j)
                j+=1
                continue 

        if j%2 or total%2:
            c=a-1 
            d=b
            if c>=0:
               x=min(total//2,c)+min(total//2,c)
               y=abs((total//2)-c) 
               #print(x,y,j,"first",c,d)
               if min(x,y)<=j and max(x,y)>=j:
                   ans.append(j)
                   j+=1
                   continue 
            c=a 
            d=b-1 
            if d>=0:
               x=min(total//2,c)+min(total//2,d)
               y=abs((total//2)-c)
               #print(x,y,j,"second",c,d) 
               if min(x,y)<=j and max(x,y)>=j:
                   ans.append(j)
            j+=1
            continue 


        if total%2==0 and j%2==0:
            x=min(total//2,a)+min(total//2,b)
            y=abs((total//2)-a) 
            if min(x,y)<=j and max(x,y)>=j:
                ans.append(j)
            j+=1
            continue 
        
        j+=1 
    print(len(ans))
    print(*ans)
        