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.
u,v=map(int,input().split())
if u>v:
    print(-1)
    exit(0)
if u==0 and v==0:
    print(0)
    exit(0)
if u==v:
    print(1)
    print(u)
    exit(0)

binary=[2**i for i in range(61)]
p=[0]*60
d=v-u
v1=v-u
u1=u
for i in range(60-1,-1,-1):
    p[i]+=u1//binary[i]
    u1%=binary[i]
answer=[]
for i in range(60-1,0,-1):
    if v1>=binary[i]:
        p[i-1]+=2
        v1%=binary[i]
f=True
if p[0]%2==1 and d%2==1:
    print(-1)
    exit(0)
m=  max(p)
while f:
    f=False
    k=0
    for i in range(60-1,-1,-1):
        if p[i]>0:
            f=True
            p[i]-=1
            k+=binary[i]
    if f:
        answer.append(k)

if sum(answer)!=v:
    print(-1)
else:
    print(m)
    print(*answer)
