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 math
from itertools import permutations


t = int(input())
for i in range(t):
    a, b = [int(x) for x in input().split()]
    posibles = set()
    p = []
    for i in range(a):
        p.append(1)
    for i in range(b):
        p.append(0)
    if abs(a - b) == 1:
        for i in range(a+b+1):
            posibles.add(i)
    elif a == b:
        for i in range(0,a+b+1, 2):
            posibles.add(i)
    elif (a % 2 == 1 and b % 2 == 1)or(a % 2 == 0 and b % 2 == 0):
        n = 0
        tmp = 0
        if (a > b):
            n = (a - b)//2
            tmp = b//2+1
        else:
            n = (b - a)//2
            tmp = a//2+1
        tmp = tmp*2
        counter = 0 + (a % 2 == 0 and b % 2 == 0)
        for i in range(n, a+b, 2):
            posibles.add(i)
            counter += 1
            if counter == tmp:
                break 
    else:
        n = 0
        small = 0
        tmp = 0
        if (a > b):
            n = (a - b)//2
            tmp = b//2+1
            small = b
        else:
            n = (b - a)//2
            tmp = a//2+1
            small = a
        counter = 0
        for i in range(n, a+b):
            posibles.add(i)
            counter+= 1
            if counter >= (small+1)*2:
                break
        
    print(len(posibles))
    s = ""
    p = sorted(posibles)
    for i in range(len(p)-1):
        print(p[i], end=" ")
    print(p[len(p)-1])