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())
ans = []
tmp = v
# xor, sum
for i in range(1000):
    if (u >> i) & 1 == 1 and (v >> i) & 1 == 0:
        if i - 1 < 0:
            print(-1)
            exit()
        ans.append((1 << i - 1))
        ans.append((1 << i - 1))
        ans.append((1 << i))
        v -= (1 << i + 1)
    elif (u >> i) & 1 == 0 and (v >> i) & 1 == 1:
        if i - 1 < 0:
            print(-1)
            exit()
        ans.append((1 << i - 1))
        ans.append((1 << i - 1))
    elif (u >> i) & 1 == 1 and (v >> i) & 1 == 1:
        ans.append((1 << i))
    elif (u >> i) & 1 == 0 and (v >> i) & 1 == 0:
        pass

if sum(ans) != tmp:
    print(-1)
    exit()

cnt = [0] * 100
for num in ans:
    for i in range(100):
        if (1 << i) & num:
            cnt[i] += 1

res = []
for i in range(len(ans)):
    tmp = 0
    for i in range(100):
        if cnt[i] >= 1:
            tmp += (1 << i)
            cnt[i] -= 1
    if tmp > 0:
        res.append(tmp)
    else:
        break
print(len(res))
print(*res)

