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.


"""
NTC here
"""
import sys
inp = sys.stdin.readline
def input(): return inp().strip()
# flush= sys.stdout.flush
# import threading
# sys.setrecursionlimit(10**6)
# threading.stack_size(2**26)

def iin(): return int(input())


def lin(): return list(map(int, input().split()))


# range = xrange
# input = raw_input

# def BFS(s, adj, step):
#     parent = {s: None}
#     ans = {}
#     ch = 0
#     u = [s]
#     while u:  # runs till u is []
#         nextu = []
#         for i in u:
#             for v in adj[i]:
#                 if v not in parent:
#                     ans[(v, i)]=step[ch]
#                     ch+=1
#                     parent[v] = i
#                     nextu.append(v)
#         u = nextu.copy()
#     return ans
def main():
    T = 1
    while T:
        T-=1
        u, v = lin()
        if u>v:print(-1)
        else:
            a = []
            ch = 0
            v -= u
            while u:
                if u&1:
                    a.append(1<<ch)
                ch+=1
                u>>=1
            if v%2==0:
                if v:
                    a.append(v//2)
                    a.append(v//2)
                chg = 1
                while chg:
                    chg = 0
                    l = len(a)
                    na = []
                    done = set()
                    for i in range(l):
                        if i in done:
                            continue
                        for j in range(i+1, l):
                            if j in done:
                                continue
                            else:
                                if a[i]&a[j]==0:
                                    chg = 1
                                    na.append(a[i]+a[j])
                                    done.update([i, j])
                                    break
                    for i in range(l):
                        if i not in done:
                            na.append(a[i])
                    # print(a, na, done)
                    a = na
                print(len(a))
                print(*a)
            else:
                print(-1)

        


            
            

main()

# threading.Thread(target=main).start()


"""
1
5
1 3
1 4
1 6
2 6
0 5
"""