XOR Break --- Solo Version
This is the solo version of the problem. Note that the solution of this problem may or may not share ideas with the solution of the game version. You can solve and get points for both versions independently.You can make hacks only if both versions of the problem are solved.Given an integer variable $$$x$$$ with the initial value of $$$n$$$. A single break operation consists of the following steps:   Choose a value $$$y$$$ such that $$$0 \lt y \lt x$$$ and $$$0 \lt (x \oplus y) \lt x$$$.  Update $$$x$$$ by either setting $$$x = y$$$ or setting $$$x = x \oplus y$$$.  Determine whether it is possible to transform $$$x$$$ into $$$m$$$ using a maximum of $$$63$$$ break operations. If it is, provide the sequence of operations required to achieve $$$x = m$$$.You don't need to minimize the number of operations.Here $$$\oplus$$$ denotes the bitwise XOR operation.
In the first test case $$$n = 7$$$, for the first operation $$$x = 7$$$ if we choose $$$y = 3$$$ then $$$(7 \oplus 3) \lt 7$$$, hence we can update $$$x$$$ with $$$3$$$ which is equal to $$$m$$$.In the second test case $$$n = 4$$$, for the first operation $$$x = 4$$$.If we choose:   $$$y = 1$$$ then $$$(4 \oplus 1) \gt 4$$$  $$$y = 2$$$ then $$$(4 \oplus 2) \gt 4$$$  $$$y = 3$$$ then $$$(4 \oplus 3) \gt 4$$$ Hence we can't do the first operation and it is impossible to make $$$x = 2$$$.
The first line contains one positive integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.Each test case consists of a single line containing two integers $$$n$$$ and $$$m$$$ ($$$1 \leq m \lt n \leq 10^{18}$$$) — the initial value of $$$x$$$ and the target value of $$$x$$$.
For each test case, output your answer in the following format.If it is not possible to achieve $$$m$$$ in $$$63$$$ operations, print $$$-1$$$.Otherwise, The first line should contain $$$k$$$ ($$$1 \leq k \leq 63$$$) — where $$$k$$$ is the number of operations required.The next line should contain $$$k+1$$$ integers — the sequence where variable $$$x$$$ changes after each break operation. The $$$1$$$-st and $$$k+1$$$-th integers should be $$$n$$$ and $$$m$$$, respectively.
import sys
input = sys.stdin.readline

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

def readInt():
    return int(input())

def readInts():
    return map(int, input().split())

def readStr():
    return input().strip()

def solve():
    (n, m) = readInts()
    s = bin(n)[2:]
    t = bin(m)[2:]
    t = '0' * (len(s) - len(t)) + t
    k = len(s)
    flag = flag2 = 0
    for i in range(k):
        if not flag:
            if s[i] == t[i]:
                flag2 |= s[i] == '1'
                continue
            elif s[i] == '1':
                flag = 1
            else:
                print(-1)
                return
        elif s[i] == '1':
            nt = t[:i] + '1' * (k - i)
            if nt == t:
                print(1)
                print(n, m)
            else:
                print(2)
                print(n, int(nt, 2), m)
            return
        elif t[i] == '0':
            continue
        else:
            if flag2:
                nt = t[:i] + '1' * (k - i)
                if nt == t:
                    print(1)
                    print(n, m)
                else:
                    print(2)
                    print(n, int(nt, 2), m)
                return
            else:
                print(-1)
            return
    print(1)
    print(n, m)
    return
for _ in range(int(input())):
    solve()
import sys
input = sys.stdin.readline

def check(A):
    for i in range(1, len(A)):
        x = A[i - 1]
        y = A[i]
        if y < x and x ^ y < x:
            pass
        else:
            return False
    return True
for _ in range(int(input())):
    (n, m) = map(int, input().split())
    if m < n and n ^ m < n:
        ANS = [n, m]
    else:
        for i in range(63, -1, -1):
            if 1 << i & n != 0:
                MAX = i
                break
        ANS = [n, 1 << MAX | m, m]
    if check(ANS) == True:
        print(len(ANS) - 1)
        print(*ANS)
    else:
        print(-1)
import sys
import math
from collections import deque
import heapq
input = lambda : sys.stdin.readline().rstrip()
ii = lambda : int(input())
mi = lambda : map(int, input().split())
ti = lambda : tuple(mi())
li = lambda : list(mi())
for _ in range(ii()):
    (n, m) = mi()
    k1 = len(bin(n)) - 3
    n2 = n - (1 << k1)
    if n2 == 0:
        print(-1)
        continue
    k2 = len(bin(n2)) - 3
    k_m = len(bin(m)) - 3
    if k1 == k_m:
        print(1)
        print(n, m)
        continue
    if k_m > k2:
        print(-1)
        continue
    if k_m == k2:
        print(1)
        print(n, m)
        continue
    print(2)
    print(n, (1 << k1) + (1 << k2) - 1, m)