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 toBinary(n):
    arr = []
    k = n
    i = 0
    while(k > 0):
        next = (k % (2 * 2 ** i)) // (2 ** i)
        arr.append(next)
        k -= next * 2 ** i
        i += 1
    return arr
fptr = sys.stdout
uv = input().split()
u = int(uv[0])
v = int(uv[1])
def f(u, v): #u <= v
    if(u > v):
        fptr.write(str(-1))
        return
    if(u == v and u == 0):
        fptr.write(str(0))
        return
    if(u == v):
        fptr.write(str(1) + '\n')
        fptr.write(str(u))
        return
    if((v - u) % 2 == 1):
        fptr.write(str(-1))
        return
    diff = (v - u) // 2
    binV = toBinary(v)
    if((u + diff) ^ diff != u):
        fptr.write(str(3) + '\n')
        fptr.write(str(u) + ' ' + str(diff) + ' ' + str(diff))
        return
    fptr.write(str(2) + '\n')
    fptr.write(str(u + diff) + ' ' + str(diff))
f(u, v)
fptr.close()