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 <bits/stdc++.h>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#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 srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto& (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define sar(a,n) {cout<<#a<<":";rep(pachico,n)cout<<" "<<a[pachico];cout<<endl;}

using namespace std;

template<typename S,typename T>auto&operator<<(ostream&o,pair<S,T>p){return o<<"{"<<p.fi<<","<<p.se<<"}";}
template<typename T>auto&operator<<(ostream&o,set<T>s){for(auto&e:s)o<<e<<" ";return o;}
template<typename S,typename T,typename U>
auto&operator<<(ostream&o,priority_queue<S,T,U>q){while(!q.empty())o<<q.top()<<" ",q.pop();return o;}
template<typename K,typename T>auto&operator<<(ostream&o,map<K,T>&m){for(auto&e:m)o<<e<<" ";return o;}
template<typename T>auto&operator<<(ostream&o,vector<T>v){for(auto&e:v)o<<e<<" ";return o;}
void ashow(){cout<<endl;}template<typename T,typename...A>void ashow(T t,A...a){cout<<t<<" ";ashow(a...);}
template<typename S,typename T,typename U>
struct TRI{S fi;T se;U th;TRI(){}TRI(S f,T s,U t):fi(f),se(s),th(t){}
bool operator<(const TRI&_)const{return(fi==_.fi)?((se==_.se)?(th<_.th):(se<_.se)):(fi<_.fi);}};
template<typename S,typename T,typename U>
auto&operator<<(ostream&o,TRI<S,T,U>&t){return o<<"{"<<t.fi<<","<<t.se<<","<<t.th<<"}";}

typedef pair<int, int> P;
typedef pair<ll, ll> pll;
typedef TRI<int, int, int> tri;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef vector<P> vp;
typedef vector<double> vd;
typedef vector<string> vs;

const int MAX_N = 100005;

ll two[61];

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n;
    cin >> n;
    two[0] = 1;
    rep(i,60){
        two[i+1] = two[i] * 2;
    }
    vi vec;
    int id = 1;
    const int sm = 2;
    vi val, tmp;
    while(n){
        int l = 1, r = 60;
        while(r-l>1){
            int mid = (l+r)/2;
            if(n >= two[mid-1]-1){
                l = mid;
            }else{
                r = mid;
            }
        }
        n -= two[l-1] - 1;
        val.pb(l);
        rep(j,l){
            tmp.pb(id);
        }
        if(id % sm == 0){
            int cur = 0;
            long long sub = 0;
            rep(i,sm){
                cur += val[i] / 2;
                sub += two[val[i] / 2] - 1;
            }
            if(n >= two[cur] - 1 - sub){
                rep(i,sm){
                    rep(j,val[i]/2){
                        vec.pb(id-i);
                    }
                }
                rep(i,sm){
                    rep(j,val[i]/2){
                        vec.pb(id-i);
                    }
                }
                rep(i,sm){
                    rep(j,val[i]%2){
                        vec.pb(id-i);
                    }
                }
                n -= two[cur] - 1 - sub;
            }else{
                rep(i,sm){
                    rep(j,val[i]){
                        vec.pb(id-i);
                    }
                }
            }
            val.clear();
            tmp.clear();
        }
        ++id;
    }
    rep(i,len(tmp)){
        vec.pb(tmp[i]);
    }
    cout << len(vec) << "\n";
    rep(i,len(vec)){
        cout << vec[i] << " ";
    }
    cout << "\n";
    return 0;
}