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.
def compute(S, X):
    A = (S - X)//2
    a = 0
    b = 0
    for i in range(64):
        Xi = (X & (1 << i))
        Ai = (A & (1 << i))
        if (Xi == 0 and Ai == 0):
            pass

        elif (Xi == 0 and Ai > 0):
            a = ((1 << i) | a)
            b = ((1 << i) | b)

        elif (Xi > 0 and Ai == 0):
            a = ((1 << i) | a)

        else:
            return -1
    if a + b != S:
        return -1
    return a, b


a, b = list(map(int, input().strip().split()))

if a == b == 0:
    print(0)
    exit()

if a == b:
    print(1)
    print(a)
    exit()

if (b-a) % 2 != 0:
    print(-1)
    exit()

o = compute(b, a)
if o != -1:
    print(2)
    print(*o)
    exit()

v = (b-a)//2

if v < 0:
    print(-1)
    exit()

print(3)
print(a, v, v)
