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 <bits/stdc++.h>
using namespace std;
//types
typedef long long ll;
typedef pair<int,int> pii;
//input
bool SR(int &_x){return scanf("%d",&_x)==1;}bool SR(ll &_x){return scanf("%lld",&_x)==1;}
bool SR(double &_x){return scanf("%lf",&_x)==1;}bool SR(char *_s){return scanf("%s",_s)==1;}
bool RI(){return true;}
template<typename I,typename... T>bool RI(I &_x,T&... _tail){return SR(_x) && RI(_tail...);}
//output
void SP(const int _x){printf("%d",_x);}void SP(const ll _x){printf("%lld",_x);}
void SP(const double _x){printf("%.16lf",_x);}void SP(const char *s){printf("%s",s);}
void PL(){puts("");}
template<typename I,typename... T>void PL(const I _x,const T... _tail)
{SP(_x);if(sizeof...(_tail)) putchar(' ');PL(_tail...);}
//macro
#define SZ(x) ((int)(x).size())
#define ALL(x) (x).begin(),(x).end()
#define REP(i,n) for(int i=0;i<int(n);i++)
#define REP1(i,a,b) for(int i=(a);i<=int(b);i++)
#define PER1(i,a,b) for(int i=(a);i>=int(b);i--)
#define pb push_back
#define mkp make_pair
#define F first
#define S second
//debug
#ifdef darry140
template<typename A,typename B>
ostream& operator <<(ostream&_s, const pair<A,B> &_p){return _s<<"("<<_p.F<<","<<_p.S<<")";}
template<typename It>
ostream& _OUTC(ostream &_s,It _b,It _e)//container
{
    _s<<"{";
    for(auto _it=_b;_it!=_e;_it++) _s<<(_it==_b?"":" ")<<*_it;
    _s<<"}";
    return _s;
}
template<typename A,typename B>
ostream& operator <<(ostream&_s, const map<A,B> &_c){return _OUTC(_s,ALL(_c));}
template<typename T>
ostream& operator <<(ostream&_s, const set<T> &_c){return _OUTC(_s,ALL(_c));}
template<typename T>
ostream& operator <<(ostream&_s, const vector<T> &_c){return _OUTC(_s,ALL(_c));}
template<typename I>
void _DOING(const char *_s,I&& _x){cerr<<_s<<"="<<_x<<endl;}//without ','
template<typename I,typename... T>
void _DOING(const char *_s,I&& _x,T&&... _tail)//with ','
{
    int _c=0;
    static const char _bra[]="({[";
    static const char _ket[]=")}]";
    while(*_s!=',' || _c!=0)//eg. mkp(a,b)
    {
        if(strchr(_bra,*_s)) _c++;
        if(strchr(_ket,*_s)) _c--;
        cerr<<*_s++;
    }
    cerr<<"="<<_x<<", ";
    _DOING(_s+1,_tail...);
}
#define debug(...) do{\
    fprintf(stderr,"%s:%d - ",__PRETTY_FUNCTION__,__LINE__);\
    _DOING(#__VA_ARGS__,__VA_ARGS__);\
}while(0)
#else
#define debug(...)
#endif
//}}}
pair<vector<int>,vector<int>> doo(ll n)
{
    if(n==1) return {vector<int>(),vector<int>()};
    vector<int> a,b;
    tie(a,b) = doo(n/2);

    assert(SZ(a)==SZ(b));
    REP(i,SZ(a)) a[i]++,b[i]++;
    a.pb(1);b.pb(1);
    if(n&1)
    {
        REP(i,SZ(a)) a[i]++,b[i]++;
        a.insert(a.begin(),1);
        b.pb(1);
    }
    return {a,b};
}
int main()
{
    ll n;RI(n);
    vector<int> a,b;
    tie(a,b) = doo(n+1);
    a.insert(a.end(),ALL(b));
    PL(SZ(a));
    REP(i,SZ(a)) printf("%d%c",a[i]," \n"[i==SZ(a)-1]);
    return 0;
}




































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