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 PYTHON3):

Solution:
x,s = map(int,input().split())
def odw(l):
	wyn = 0
	pot = 1
	for i in l:
		wyn += pot * i
		pot *= 2
	return wyn
def konw(n):
	l = []
	while n > 0:
		l.append(n%2)
		n //= 2
	return l
xx = konw(x)
ss = konw(s)
if x == 0 and s == 0:
	print(0)
else:
	if x > s:
		print(-1)
	if x == s:
		print(1)
		print(x)
	if x < s:
		if (x-s)%2 == 1:
			print(-1)
		else:
			kan1 = []
			kan2 = []
			for i in range(len(xx)):
				if xx[i] == 0:
					kan1.append(0)
					kan2.append(0)
				else:
					kan1.append(1)
					kan2.append(0)
			for j in range(len(xx),len(ss)):
				kan1.append(0)
				kan2.append(0)
			ind = len(ss)-1
			while odw(kan1) + odw(kan2) < s:
				while ind >= 0:
					if not (kan1[ind] == kan2[ind] and kan1[ind] == 0):
						ind -= 1
					else:
						if 2**(ind+1)+odw(kan1)+odw(kan2) > s:
							ind -= 1
						else:
							break
				if ind == -1:
					break
				kan1[ind] = 1
				kan2[ind] = 1
			if odw(kan1)+odw(kan2) == s:
				print(2)
				print(odw(kan1),odw(kan2))
			else:
				print(3)
				print(x, (s-x)//2,(s-x)//2)