Problem: Given 2 integers u and v, find the shortest array such that [bitwise-xor](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of its elements is u, and the sum of its elements is v.

Input

The only line contains 2 integers u and v (0 ≤ u,v ≤ 10^{18}).

Output

If there's no array that satisfies the condition, print "-1". Otherwise:

The first line should contain one integer, n, representing the length of the desired array. The next line should contain n positive integers, the array itself. If there are multiple possible answers, print any.

Examples

Input


2 4


Output


2
3 1

Input


1 3


Output


3
1 1 1

Input


8 5


Output


-1

Input


0 0


Output


0

Note

In the first sample, 3⊕ 1 = 2 and 3 + 1 = 4. There is no valid array of smaller length.

Notice that in the fourth sample the array is empty.

Solution (in PYTHON):

Solution:
from sys import stdin, stdout
from collections import Counter, defaultdict


# Fast IO utilities
raw_input = stdin.readline
pr = stdout.write


def in_arr():
    return map(int,raw_input().split())


def pr_num(n):
    stdout.write(str(n)+'\n')


def pr_arr(arr):
    for i in arr:
        stdout.write(str(i)+' ')
    stdout.write('\n')

# commment out below line for python version 3.0+
range = xrange

# main code

u,v=in_arr()
dp1=[0]*65
dp2=[0]*65
for i in range(62):
    val=1<<i
    if val&u:
        dp2[i]+=1
    if val&v:
        dp1[i]+=1
ans=[0]*65
for i in range(61,-1,-1):
    if dp2[i]:
        if not dp1[i]:
            print -1
            exit()
        ans[i]=1
        dp1[i]-=1
        if i:
            dp1[i-1]+=2*dp1[i]
    elif dp1[i]:
        if i:
            dp1[i-1]+=2*dp1[i]
            dp1[i]-=1
        elif dp1[i]%2:
            print -1
            exit()
if dp1[0]%2:
    print -1
    exit()
val=dp1[0]
val/=2
for i in range(62):
    if (1<<i)&val:
        ans[i]+=2
ans1=max(ans)
arr=[0]*ans1
for i in range(62):
    val=1<<i
    for j in range(ans[i]):
        arr[j]+=val
print ans1
for i in arr:
    print i,
print