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 collections import *
from functools import lru_cache
import bisect
inf = float('inf')
ip = input


def ipl():
    return list(map(int, input().split()))


def ipn():
    return int(input())


def ipf():
    return float(input())


def solve():
    a, b = ipl()
    n = a + b
    k = set()
    for i in range(min(a, b) + 1):
        if n % 2 == 0:
            r = (n // 2) - min(a, b) + (i * 2)
            k.add(r)
        else:
            k.add((n // 2) - min(a, b) + (i * 2))
            k.add((n // 2) - min(a, b) + (i * 2) + 1)

    print(len(k))
    for i in sorted(list(k)):
        print(i, end=" ")
    print()


for _ in range(int(input())):
    solve()
