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. 
def subsets_gen(num):
    def generator_h(no, s):
        if (len(s) == no):
            yield s
        else:
            yield from generator_h(no, s+"0")
            yield from generator_h(no, s+"1")
    yield from generator_h(num, "")

def perms_gen(pole):
    used = [False for i in range(len(pole))]
    def generator_h(l, m):
        if len(m) == len(l):
            yield m
        else:
            for i in range(len(l)):
                if used[i]:
                    continue
                else:
                    m.append(l[i])
                    used[i] = True
                    yield from generator_h(l, m)
                    m.pop(); used[i] = False
    yield from generator_h(pole, [])

def fact(n):
    factors = []
    if (n%2 == 0):
        while(n%2 == 0):
            factors.append(2)
            n//=2
    poc = 3
    while(poc*poc<=n):
        while(n%poc == 0):
            factors.append(poc)
            n//=poc
        poc += 2
    if n>1:
        factors.append(n)
    return factors

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

def NcK(n, k):
    if (k == 0 or k == n):
        return 1
    return NcK(n-1, k-1)+NcK(n-1, k)
 
testcases = 1
testcases = int(input())
for _ in range(testcases):
    a, b = map(int, input().split())
    if (a+b)%2 == 1:
        x = (a+b+1)//2 + min(a, b)
        y = (max(a, b)-(a+b+1)//2)
        print(x-y+1)
        for i in range(y, x+1):
            if (i == x):
                print(i)
            else:
                print(i, end = " ")
    else:
        x = (a+b)//2 + min(a, b)
        y = (max(a, b)-(a+b)//2)
        print(x-y)
        for i in range(y, x+1, 2):
            if (i == x):
                print(i)
            else:
                print(i, end = " ")
