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())

def d(v = -1):
    print(v)
    exit(0)

if u > v:
    d()

if u == v:
    if u == 0:
        d(0)
    else:
        print(1)
        d(u)

# u < v
# if u % 2 != v % 2:
#     d()

bit_u = [0] * 600
bit_v = [0] * 600

for i in range(64):
    if (1 << i) & u:
        bit_u[i] = 1
    if (1 << i) & v:
        bit_v[i] = 1
for i in range(63, 0, -1):
    if bit_u[i] == bit_v[i] % 2:
        bit_v[i - 1] += (bit_v[i] >> 1) << 2
        bit_v[i] %= 2
    else:
        if bit_v[i] == 0:
            d()
        bit_v[i] -= 1
        bit_v[i - 1] += 2
        bit_v[i - 1] += (bit_v[i] >> 1) << 2
        bit_v[i] %= 2

for i in range(500):
    up = (bit_v[i] >> 2) << 1
    bit_v[i + 1] += up;
    bit_v[i] -= up << 1

n = max(bit_v)
res = [0] * n
for i in range(n):
    for j in range(600):
        if bit_v[j] > i:
            res[i] |= 1 << j

xr, sm = 0, 0
for i in res:
    xr ^= i
    sm += i
if xr != u or sm != v:
    d()
    exit(0)

print(n)
print(*res)
