Problem: Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.

Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.

In the Cat Furrier Transform, the following operations can be performed on x:

  * (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
  * (Operation B): replace x with x + 1. 



The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.

Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?

Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.

Input

The only line contains a single integer x (1 ≤ x ≤ 10^6).

Output

The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.

Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.

If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.

Examples

Input


39


Output


4
5 3 

Input


1


Output


0


Input


7


Output


0

Note

In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:

  1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56. 
  2. Increase x by 1, changing its value to 57. 
  3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62. 
  4. Increase x by 1, changing its value to 63 = 2^6 - 1. 



In the second and third test, the number already satisfies the goal requirement.

Solution (in PYTHON):

Solution:
"""
Author    : raj1307 - Raj Singh
Institute : Jalpaiguri Government Engineering College
Date      : 24.04.19
"""
from __future__ import division, print_function
import itertools,os,sys
from collections import deque, Counter, OrderedDict, defaultdict
#from heapq import nsmallest, nlargest, heapify, #heappop ,heappush, heapreplace
#from math import ceil,floor,log,sqrt,factorial,pow,pi
#from bisect import bisect_left,bisect_right
#from decimal import *

import os
import sys
from io import BytesIO, IOBase

if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
else:
    from builtins import str as __str__
    str = lambda x=b'': x if type(x) is bytes else __str__(x).encode()

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._buffer = BytesIO()
        self._fd = file.fileno()
        self._writable = 'x' in file.mode or 'r' not in file.mode
        self.write = self._buffer.write if self._writable else None

    def read(self):
        return self._buffer.read() if self._buffer.tell() else os.read(self._fd, os.fstat(self._fd).st_size)

    def readline(self):
        while self.newlines == 0:
            b, ptr = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)), self._buffer.tell()
            self._buffer.seek(0, 2), self._buffer.write(b), self._buffer.seek(ptr)
            self.newlines += b.count(b'\n') + (not b)
        self.newlines -= 1
        return self._buffer.readline()

    def flush(self):
        if self._writable:
            os.write(self._fd, self._buffer.getvalue())
            self._buffer.truncate(0), self._buffer.seek(0)


sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
input = lambda: sys.stdin.readline().rstrip(b'\r\n')


def print(*args, **kwargs):
    sep, file = kwargs.pop('sep', b' '), kwargs.pop('file', sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop('end', b'\n'))
    if kwargs.pop('flush', False):
        file.flush()





def ii(): return int(input())
def si(): return str(input())
def mi():return map(int,input().strip().split(" "))
def li():return list(mi())

abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
def getKey(item): return item[0] 
def sort2(l):return sorted(l, key=getKey)
def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
def ntl(n):return [int(i) for i in str(n)]
def powerMod(x,y,p):
    res = 1
    x %= p
    while y > 0:
        if y&1:
            res = (res*x)%p
        y = y>>1
        x = (x*x)%p1
    return res
def gcd(x, y):
    while y:
        x, y = y, x % y
    return x

# For getting input from input.txt file 
#sys.stdin = open('input.txt', 'r')  
  
# Printing the Output to output.txt file 
#sys.stdout = open('output.txt', 'w') 

def isPowerOfTwo (x): 
    return (x and (not(x & (x - 1))) )

def decimalToBinary(n): 
    return bin(n).replace("0b","")

def main():
   
    
    n=ii()
    c=0
    if isPowerOfTwo(n+1):
        print(0)
        exit()
    m=[]
    op=0
    
    l=list(decimalToBinary(n))
    num=0
    
        
    
    
    
    
    while(True):
        if isPowerOfTwo(n+1):
            break
        else:
            if c%2:
                n+=1
                op+=1
                c+=1
            else:
                l=list(decimalToBinary(n))
                le=len(l)
                #print(l)
                if l.count('0')!=0:
                    p=l.index('0')
                    le-=p
                    n^=((2**le)-1)
                    #print(le)
                    #print((2^(2**le)-1))
                else:
                    le=0
           
                m.append(le)
                op+=1
                c+=1
                
    print(op)
    print(*m)
        
    
    
    
    
        
   


if __name__ == "__main__":
    main()