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.
from sys import stdin
# input=stdin.buffer.readline
input=lambda : stdin.readline().strip()
lin=lambda :list(map(int,input().split()))
iin=lambda :int(input())
main=lambda :map(int,input().split())
from math import ceil,sqrt,factorial,log
from collections import deque
from bisect import bisect_left
mod=998244353
mod=1000000007
def gcd(a,b):
	a,b=max(a,b),min(a,b)
	while a%b!=0:
		a,b=b,a%b
	return b
def moduloinverse(a):
	return(pow(a,mod-2,mod))
def solve(we):
	a,b=main()
	if a>b or a%2!=b%2:
		print(-1)
		return
	if a==b:
		if a==0:
			print(0)
		else:
			print(1)
			print(a)
		return
	x=(b-a)//2
	if a&x:
		print(3)
		print(a,x,x)
	else:
		print(2)
		print(a^x,x)




qwe=1
# qwe=iin()
for _ in range(qwe):
	solve(_+1)