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.
"""
    Template written to be used by Python Programmers.
    Use at your own risk!!!!
    Owned by enraged(rating - 5 star at CodeChef and Specialist at Codeforces).
"""
import sys
import heapq
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt
from collections import defaultdict as dd, deque, Counter as c
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
# sys.setrecursionlimit(2*pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(str(var))
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [[val for i in range(n)] for j in range(m)]


def two(n):
    if n == 0:
        return True
    i = 0
    while i <= 64:
        if n == pow(i, 2):
            return True
        i += 1
    return False


def factors(n):
    s = set()
    i = 1
    while i <= sqrt(n):
        if n % i == 0:
            s.add(i)
            s.add(n//i)
        i += 1
    return s


u, v = sp()
if u + v == 0:
    out(0)
    exit()
if u > v:
    out(-1)
    exit()
if (u % 2) + (v % 2) == 1:
    out(-1)
    exit()
if u == v:
    out("1\n"+str(u))
    exit()
x = (v-u)//2
if u & x:
    out('3\n')
    out(str(u)+" "+str(x)+" "+str(x))
    exit()
out('2\n')
out(str(u+x)+" "+str(x))
