Problem:
Score : 1000 points

Problem Statement

We will call a string x good if it satisfies the following condition:

Condition: x can be represented as a concatenation of two copies of another string y of length at least 1.

For example, aa and bubobubo are good; an empty string, a, abcabcabc and abba are not good.

Eagle and Owl created a puzzle on good strings.
Find one string s that satisfies the following conditions. It can be proved that such a string always exists under the constraints in this problem.

1 ≤ |s| ≤ 200

Each character of s is one of the 100 characters represented by the integers 1 through 100.

Among the 2^{|s|} subsequences of s, exactly N are good strings.

Constraints

1 ≤ N ≤ 10^{12}

Input

Input is given from Standard Input in the following format:

N

Output

In the first line, print |s|, the length of s.
In the second line, print the elements in s in order, with spaces in between. Any string that satisfies the above conditions will be accepted.

Sample Input 1

7

Sample Output 1

4
1 1 1 1

There are two good strings that appear as subsequences of s: (1,1) and (1,1,1,1). There are six occurrences of (1,1) and one occurrence of (1,1,1,1), for a total of seven.

Sample Input 2

299

Sample Output 2

23
32 11 11 73 45 8 11 83 83 8 45 32 32 10 100 73 32 83 45 73 32 11 10

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<queue>
#define LL long long
using namespace std;
inline LL read( )
{
  LL sum=0;char c=getchar( );bool f=0;
  while(c<'0'||c>'9') {if(c=='-') f=1;c=getchar( );}
  while(c>='0'&&c<='9') {sum=sum*10+c-'0';c=getchar( );}
  if(f) return -sum;
  return sum;
}
LL n;int len;bool p[202];
int main( )
{
  n=read( )+1;
  while(n!=1)
    {
      if(n&1) n--,p[++len]=1;
      else n>>=1,p[++len]=0;
    }
  printf("%d\n",len<<1);
  int i;deque<int>q;
  for(i=len;i>=1;i--)
    p[i]?q.push_front(len-i+1):q.push_back(len-i+1);
  while(!q.empty( )) 
    printf("%d ",q.front( )),q.pop_front( );
  for(i=1;i<=len;i++) printf("%d ",i);
  return 0;
}