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 

u, v = map(int, stdin.readline().split())


k = u

if (u - v) % 2 != 0:
	#
	print (-1)
	exit()
if u == v:
	if u == 0:
		print (0)
		exit()
	else:
		print (1)
		print (u)
		exit()

if v < u:
	print (-1)
	exit()

# 2 or 3 
x = (v - u)//2 
u = [int(z) for z in list(bin(u)[2:][::-1])]
x_b = [int(z) for z in list(bin(x)[2:][::-1])]
l = max(len(u), len(x_b))
u += [0]*(l - len(u))
x_b += [0]*(l - len(x_b))

ans = []
c, d = [], []
for i in range(l):
	a, b = u[i], x_b[i]
	if a == b == 1:
		print (3)
		print (k, x, x)
		break
	else:
		if a == b == 0:
			c.append(0)
			d.append(0)
		elif a == 1 and b == 0:
			c.append(1)
			d.append(0)

		else:
			c.append(1)
			d.append(1)


	if i == l - 1:
		c = ''.join([str(x) for x in c[::-1]])
		d = ''.join([str(x) for x in d[::-1]])
		c, d = int(c, 2), int(d, 2)
		print(2)
		print(c, d)






