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:
from collections import Counter
from collections import deque
from sys import stdin
from bisect import *
from heapq import *
import math

g   = lambda : stdin.readline().strip()
gl  = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")

x, s = gil()
if x > s or (s-x)&1:
	print(-1)
else:
	d = (s-x)//2
	a = [x, d, d]
	# print(math.log2(s))
	for i in range(1, 3):
		for j in range(60):
			bit = (1<<j)
			if (a[i] & bit) and (a[i-1]&bit)==0:
				# print(j, "th bit toggling inn ", a[i-1], a[i])
				a[i] = a[i]^bit;a[i-1] = a[i-1]^bit
				# print(j, "th bit toggling inn: after ", a[i-1], a[i])

	while a and a[-1]==0:
		a.pop()
	print(len(a))
	if a:print(*a)
	# print(sum(a)==s)