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.
import sys
def fastio():
    from io import StringIO
    from atexit import register
    global input
    sys.stdin = StringIO(sys.stdin.read())
    input = lambda : sys.stdin.readline().rstrip('\r\n')
    sys.stdout = StringIO()
    register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()
 
INF = 10**20
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
from math import gcd
from math import ceil
from collections import defaultdict as dd, Counter
from bisect import bisect_left as bl, bisect_right as br
 
u, v = I()
 
if u > v:
    print(-1)
    exit()
 
if u == v and u:
    print(1)
    print(u)
    exit()
if u == v and not u:
    print(0)
    exit()
ans = []
cu, cv = u, v
 
def mask(x, bit):
    return (x >> bit) & 1
 
d = dd(int)
v -= u
 
if v % 2:
    print(-1)
    exit()
# ans.append(u)
for i in range(65, -1, -1):
    if mask(u, i):
        ans.append(2 ** i)
for bit in range(65, 0, -1):
    if mask(v, bit):
        value = 2 ** (bit - 1)
        d[bit - 1] = 1
        ans.append(value)
        ans.append(value)
res = []
while ans:
    sum = 0
    for i in set(ans):
        ans.remove(i)
        sum += i
    res.append(sum)
print(len(res))
print(*res)