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.


x,s=map(int,input().split())

if(x==0 and s==0):
	print(0)
	exit()
if(x>s):
	print(-1)
	exit()
if(x&1!=s&1):
	print(-1)
	exit()
if(x==s):
	print(1)
	print(x)
	exit()
if(x==0):
	print(2)
	print(s//2,s//2)
	exit()
b=(s-x)//2

f=1

bx=bin(x)[2:]
ba=""
bb=bin(b)[2:]

while(len(bb)<len(bx)):
	bb="0"+bb
while(len(bx)<len(bb)):
	bx="0"+bx

for i in range(len(bx)):
	if(bb[i]=="1"):
		ba+="1"
	elif(bb[i]=="0" and bx[i]=="1"):
		ba+="1"
	else:
		ba+="0"
a=int(ba,2)

if((a^b)==x and (a+b)==s):
	print(2)
	print(a,b)
	exit()

print(3)
print(x,(s-x)//2,(s-x)//2)
