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.
MX = 64
u, v = map(int, input().split())
cx = list(map(int, reversed(bin(u)[2:])))
cx += [0] * (MX - len(cx))
cs = list(map(int, reversed(bin(v)[2:])))
cs += [0] * (MX - len(cs))

imp = False
for i in reversed(range(MX)):
    if cs[i] % 2 == cx[i]:
        continue
    if i == 0:
        imp = True
        break
    if cx[i] and cs[i] == 0:
        for j in range(i + 1, MX):
            if cs[j] >= 2:
                break
        else:
            imp = True
            break
        cs[j] -= 2
        for k in range(i + 1, j):
            cs[k] += 2
        cs[i] += 4
    cs[i] -= 1
    cs[i - 1] += 2

if imp:
    print(-1)
else:
    ans = [0] * max(cs)
    for i in range(MX):
        for j in range(cs[i]):
            ans[j] += 2 ** i
    print(len(ans))
    print(*ans)
