from collections import defaultdict, deque, Counter
from heapq import heapify, heappop, heappush
import math
from copy import deepcopy
from itertools import combinations, permutations, product, combinations_with_replacement
from bisect import bisect_left, bisect_right

import sys

def input():
    return sys.stdin.readline().rstrip()
def getN():
    return int(input())
def getNM():
    return map(int, input().split())
def getList():
    return list(map(int, input().split()))
def getListGraph():
    return list(map(lambda x:int(x) - 1, input().split()))
def getArray(intn):
    return [int(input()) for i in range(intn)]

mod = 10 ** 9 + 7
MOD = 998244353
sys.setrecursionlimit(10000000)
# import pypyjit
# pypyjit.set_param('max_unroll_recursion=-1')
inf = float('inf')
eps = 10 ** (-15)
dy = [0, 1, 0, -1]
dx = [1, 0, -1, 0]

#############
# Main Code #
#############

def make_divisors(n):
    divisors = []
    for i in range(1, int(math.sqrt(n)) + 1):
        if n % i == 0:
            divisors.append(i)
            # √nで無い数についてもう一個プラス
            if i != n // i:
                divisors.append(n // i)
    return sorted(divisors)

"""
素数のみで基底を作る
2^N以上の素数

約数の総和で基底を組む　それぞれ線形独立になるよう
2^Nを基底にすればいい　約数の総和は2^(N+1)-1
"""

# 基底を40個
L = [2 ** i for i in range(40)]
vec = [2 ** (i + 1) - 1 for i in range(40)]
N = getN()
ans = []
for i in range(39, -1, -1):
    if N & (1 << i):
        ans.append(L[i])
        N ^= vec[i]

if len(ans) == 0:
    print(-1)
else:
    print(len(ans))
    print(*ans[::-1])