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 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()
        n = a+b
        ans = []
        for i in range(0,n+1):
            # A => start first
            if n % 2 == 0:
                wants_order = n-i
                max_possible = min(a,n//2)+min(b,n//2)
                min_possible = max(0,a-n//2)+max(0,b-n//2)
                # print(min_possible,max_possible)
                if wants_order <= max_possible and min_possible <= wants_order:
                    if i % 2 == 1:
                        max_not_in_order_a = min(a,n//2,i)
                        in_order_a = a - max_not_in_order_a
                        max_not_in_order_b = (n//2 - in_order_a)
                        in_order_b = b - max_not_in_order_b

                        if in_order_a + in_order_b == i:
                            ans.append(i)
                        else:
                            max_not_in_order_b = min(b, n // 2, i)
                            in_order_b = b - max_not_in_order_b
                            max_not_in_order_a = (n//2 - in_order_b)
                            in_order_a = a - max_not_in_order_a

                            if in_order_a + in_order_b == i:
                                ans.append(i)

                    else:
                        ans.append(i)
            else:
                wants_order = n-i
                possible1 = min(a,n//2)+min(b,n-n//2)
                min_possible1 = max(0, a - n // 2) + max(0, b -(n - n // 2))
                possible2 = min(a,n-n//2)+min(b,n//2)
                min_possible2 = max(0, a -(n - n // 2)) + max(0, b - n // 2)
                if (wants_order <= possible1 and min_possible1 <= wants_order) or (wants_order <= possible2 and min_possible2 <= wants_order):
                    ans.append(i)

        print(len(ans))
        print(*ans)



if __name__ == '__main__':
    main()