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. 
from itertools import combinations, permutations
from sys import path_hooks, stdin, stdout
from collections import Counter,deque
import math
from copy import deepcopy
import random
import heapq
import sys
sys.setrecursionlimit(10**5)

 
def mapinput():
    return  map(int, stdin.readline().split())

def strinput():
    return stdin.readline().strip()

def listinput():
    return list(map(int,stdin.readline().split()))

def intinput():
    return int(stdin.readline().strip())

def gcd(a,b):
    if(b==0):
        return a
    else:
        return gcd(b,a%b)

def SieveOfEratosthenes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 1
    prime[0]= False
    prime[1]= False
    ans = []
    for p in range(n + 1):
        if prime[p]:
            ans.append(p)
    return ans

coprime1 = {6: 100003, 5: 10007, 4: 1009, 3: 101, 2: 11, 1: 2, 7: 1000003, 8: 10000019, 9: 100000007}
coprime2 = {8:10000169,9:100000049,7:1000033,6:100019,5:10009,4:1013,3:103,2:13,1:3}
 
def writ(ss):
    stdout.write(str(ss) + "\n")

mod =  998244353

def sumn(n):
    return (n*(n+1))//2


def perm(lis):
    if len(lis) < 2:
        return lis
    #print(lis)
    anss = []
    for i in range(len(lis)):
        ele = lis[i]
        liss = lis[:i] + lis[i+1:]
        for j in perm(liss):
            anss.append( ele+j )
    return anss

for test in range(intinput()):
    def solve():
        #n = intinput()
        a , b = mapinput()
        if a + b & 1 == 0:
            n = (a + b) >> 1
            aa = abs(  (n & 1) - (a & 1)  )
            ans = deque([i for i in range( aa , a + b + 1 , 2 )])


        else:
            ans = deque([ i for i in range(a+b+1)])
        diff = abs( a - b) >> 2
        for i in range(diff):
            ans.popleft()
            ans.pop()
        print(len(ans))
        return ans








    print(* solve() )

