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: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <sstream>
#include <cmath>
#include <set>
#include <iomanip>
#include <deque>
#include <stdio.h>
#include <fstream>
using namespace std;

#define REP(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define RREP(i,n) for(int (i)=(int)(n)-1;i>=0;i--)
#define FILL(Itr,n) fill((Itr).begin(),(Itr).end(),n)
#define REMOVE(Itr,n) (Itr).erase(remove((Itr).begin(),(Itr).end(),n),(Itr).end())
#define UNIQUE(Itr) sort((Itr).begin(),(Itr).end()); (Itr).erase(unique((Itr).begin(),(Itr).end()),(Itr).end())
#define LBOUND(Itr,val) lower_bound((Itr).begin(),(Itr).end(),(val))
#define UBOUND(Itr,val) upper_bound((Itr).begin(),(Itr).end(),(val))
#define MOD 1000000007
typedef long long ll;
typedef pair<int,int> P;

int main(){
    
    ll N; cin>>N;
    vector<int> ans;
    REP(i,64){
        if((N>>i)&1){
            if(i==0)REP(j,2)ans.push_back(50);
            else{
                REP(j,2)ans.push_back(100-i);
                REP(j,i+1)ans.push_back(i+1);
            }
        }
    }
    
    cout<<ans.size()<<endl;
    REP(i,ans.size()-1)cout<<ans[i]<<" ";
    cout<<ans[ans.size()-1]<<endl;
    
    return 0;
}