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:
import sys
import math
def alele(): return list(map(int, sys.stdin.readline().strip().split()))
def ilele(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
import bisect
from collections import defaultdict
INF = 10 ** 18
MOD = 10 ** 9 + 7

u,v = ilele()
if v<u:
    print(-1)
elif v==u==0:
    print(0)
elif v==u:
    print(1)
    print(u)
elif v%2 != u%2: #last digit must be same
    print(-1)
else:
    y = (v-u)//2
    a = u+y;b =y
    if a+b == v and a^b == u:
        print(2)
        print(u+y,y)
    else:
        print(3)
        print(u,y,y)