Problem: 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.

Solution (in PYTHON3):

Solution:
from collections import Counter
from collections import defaultdict
import math
import random
import heapq as hq
from math import sqrt
import sys
from functools import reduce

# sys.setrecursionlimit(10000000)


def input():
    return sys.stdin.readline().strip()


def iinput():
    return int(input())


def tinput():
    return input().split()


def rinput():
    return map(int, tinput())


def rlinput():
    return list(rinput())


mod = int(1e9)+7


def factors(n):
    return set(reduce(list.__add__,
                      ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))


# ----------------------------------------------------


if __name__ == "__main__":
    u, v = rinput()
    if u == 0 and v == 0:
        print(0)
    elif u > v or (v-u) % 2:
        print(-1)
    elif u == v:
        print(1)
        print(u)
    elif (u+(v-u)//2) ^ ((v-u)//2) == u:
        print(2)
        print(u+(v-u)//2, ((v-u)//2))
    else:
        print(3)
        print((v-u)//2, (v-u)//2, u)