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.
import math
from bisect import bisect_left, bisect_right
from sys import stdin, stdout, setrecursionlimit
from collections import Counter
input = lambda: stdin.readline().strip()
print = stdout.write

u, v = map(int, input().split())
x = (v-u)//2
if (v-u)%2 or x<0:
    print('-1\n')
elif x==0:
    if u==0:
        print('0\n')
    else:
        print('1\n')
        print(str(u)+'\n')
else:
    p = bin(x)[2:]
    q = bin(u)[2:]
    p = p.rjust(max(len(p), len(q)), '0')
    q = q.rjust(max(len(p), len(q)), '0')
    a = ''
    b = ''
    for i in range(max(len(p), len(q))):
        if p[i]=='0' and q[i]=='0':
            a+='0'
            b+='0'
        elif p[i]=='0' and q[i]=='1':
            a+='0'
            b+='1'
        elif p[i]=='1' and q[i]=='0':
            a+='1'
            b+='1'
        else:
            print('3\n')
            print(str(u)+' '+str(x)+' '+str(x)+'\n')
            break
    else:
        print('2\n')
        print(str(int(a, 2))+' '+str(int(b, 2))+'\n')
