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())
if u % 2 != v % 2 or u > v:
    print(-1)
    exit()
s = bin(u)[2:].rjust(60, '0')
cnt = [0] * 60
for i in range(60):
    cnt[i] += int(s[i])
dif = v - u
for i in range(60):
    if i == 55:
        h = 0
    cur = dif // (1 << (60 - i - 1))
    cur -= cur % 2
    if (1 << (60 - i - 1)) <= dif and cur != 0:
        cnt[i] += cur
        dif -= (1 << (60 - i - 1)) * cur
#for i in range(50, 60):
#    print(cnt[i])
ans = [[0 for i in range(60)] for j in range(max(cnt))]
for i in range(len(ans)):
    for j in range(60):
        if ans[i][j] == 0 and cnt[j] != 0:
            ans[i][j] += 1
            cnt[j] -= 1
for i in range(len(ans)):
    for j in range(60):
        ans[i][j] = str(ans[i][j])
print(len(ans))
for i in ans:
    print(int(''.join(i), 2), end=' ')