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.
"""
    Author : thekushalghosh
    Team   : CodeDiggers
"""
import sys,math
input = sys.stdin.readline
############ ---- USER DEFINED INPUT FUNCTIONS ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(s[:len(s) - 1])
def invr():
    return(map(int,input().split()))
################################################################
############ ---- THE ACTUAL CODE STARTS BELOW ---- ############
t = 1
for tt in range(t):
    u,v = invr()
    if (u - v) % 2 != 0 or u > v:
        print(-1)
    elif u == 0 and v == 0:
        print(0)
    elif u == v:
        print(1)
        print(u)
    elif ((u) & ((v - u) // 2)) == 0:
        print(2)
        print((u + v) // 2,(v - u) // 2)
    else:
        print(3)
        print(u,(v - u) // 2,(v - u) // 2)