human
Write me a Python code to find all values of k such that exactly k breaks could happen during the match between Alice and Borys in total given the following:
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...
impor
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#Here is Python code to find all values of k such that exactly k breaks could happen during the match between Alice and Borys in total.
t math
from collections import deque, defaultdict
from math import sqrt, ceil, factorial, floor, inf, gcd
import bisect
import sys
import random
import heapq

def get_array(): return list(map(int, sys.stdin.readline().strip().split()))

def get_ints(): return map(int, sys.stdin.readline().strip().split())

def input(): return sys.stdin.readline().strip()

from types import GeneratorType

def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to

    return wrappedfunc


def main():
    for _ in range(int(input())):
        (a,b) = get_ints()
        a,b = max(a,b),min(a,b)
        n = a+b
        ans = []
        for i in range(a+1):
            # i represents i people present in the ath area
            if n %2 == 0:
                if i <= n//2:
                    if b >= n//2-i:
                        max_not_order_b = (n//2 - i)
                        max_not_order_a = a - i
                        ans.append(max_not_order_a+max_not_order_b)
                    else:
                        max_not_order_b = b
                        max_not_order_a = (a-i-(n//2-i-b))
                        ans.append(max_not_order_a+max_not_order_b)
            else:
                if i <= n//2:
                    if b >= n//2-i:
                        max_not_order_b = (n//2 - i)
                        max_not_order_a = a - i
                        ans.append(max_not_order_a+max_not_order_b)
                    else:
                        max_not_order_b = b
                        max_not_order_a = (a-i-((n+1)//2-i-b))
                       ...