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>
using namespace std;
#define rep(i,a,b) for(int i=a;i<b;i++)
#define iter(it,a) for(auto it=a.begin();it!=a.end();it++)
#define mp make_pair
#define pb push_back
#define F first
#define S second
// #define endl '\n'
#define __fd cout.setf(ios::fixed); cout.precision(18)
#define preturn(X) cout<<X<<endl; return 0
typedef long long ll;

vector<int> x;
map<int,int> h;

void func(){
	int y= 1;
	int i = 0;
	int j = 300;
	int j1 = 301;
	for(int k = x.size()-1;k>=0;k--){
		if(x[k]==1){
			h[j1]=y;
			j1++;
			h[j1]=y;
			j1++;
			y++;
		}
		else{
			h[i]=y;
			i++;
			h[j1]=y;
			j1++;
			y++;
		}
	}
}

int main(){
	cin.tie(0);ios_base::sync_with_stdio(0);
	ll n;
	cin>>n;
	while(n!=0){
		if(n%2==0){
			n-=(1);
			x.pb(1);
		}
		else{
			n/=2;
			x.pb(0);
		}
	}
	func();
	stringstream ss;
	int ans = 0;
	iter(it,h){
		if((*it).S != 0){
			ss<<(*it).S<<" ";	
			ans++;
		}
	}
	cout<<ans<<endl<<ss.str()<<endl;
}