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.
''' Ehab the Xorcist
'''

from math import ceil

def findpow2(integer):
    pow2 = 1
    while pow2 <= integer:
        pow2 *= 2
    return pow2

''' routine '''

U, V = list(map(int, input().split()))

carry = V - U
if carry < 0 or carry%2 == 1:
    print(-1)
elif V == 0:
    print(0)
elif carry == 0:
    print(1)
    print(U)
else:
    # 3 case
    B = carry // 2
    if (U + B)^B == U:
        nums = (U+B, B)
    else:
        nums = (U, B, B)
    qty = len(nums)
    nums = ' '.join(list(map(str, nums)))
    print(qty)
    print(nums)