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())

kouho = [u]

k = v-u
if k < 0 or k % 2 != 0:
    if u == v:
        print(1)
        print(u)
    else:
        print(-1)
    exit()

k //= 2
kouho.append(k)
kouho.append(k)
if kouho[1] == 0:
    if u == 0:
        print(0)
    else:
        print(1)
        print(u)
elif kouho[0] ^ kouho[1] == kouho[0] + kouho[1]:
    print(2)
    print(kouho[0]+kouho[1], kouho[2])
else:
    print(3)
    print(kouho[0], kouho[1], kouho[2])
