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 math import log,ceil
x,y=map(int,input().split())
bi=bin(x)
s=[]
k=y
ans,ans2=[],[]
for i in range(len(bi)-1,1,-1):
    s.append(int(bi[i]))
for i in range(65):
    s.append(0)
if y>0:
    for i in range(ceil(log(y,2))+1,-1,-1):
        if s[i] == 1:
            k -= 2 ** (i)
            ans.append(1)
            ans2.append(0)
        else:
            if k >= 2 ** (i + 1):
                k -= 2 ** (i + 1)
                ans.append(1)
                ans2.append(1)
            else:
                ans.append(0)
                ans2.append(0)
if x%2!=y%2 or x>y:
    print(-1)
elif x==y==0:
    print(0)
elif x==y:
    print(1)
    print(x)
elif x==0:
    print(2)
    print(y//2,y//2)
elif k==0:
    print(2)
    ans.reverse()
    ans2.reverse()
    a,b=0,0
    for i in range(len(ans)):
        a+=ans[i]*(2**(i))
    for i in range(len(ans)):
        b+=ans2[i]*(2**(i))
    print(a,b)
else:
    print(3)
    print(x,(y-x)//2,(y-x)//2)