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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <memory.h>
#include <queue>

#define fr first
#define sc second
#define mk make_pair

using namespace std;

long long n;
vector<int> vec;

int main()
{
    deque<int> ans;
    cin >> n;
    while(n)
    {
        if(n % 2 == 1)
        {
            vec.push_back(1);
            n = n / 2;
        }else
        {
            vec.push_back(0);
            n--;
        }
    }

    int p = 1;
    for(int i = vec.size() - 1; i >= 0; i--)
    {
        if(vec[i] == 1)
        {
            ans.push_back(p);
        }else
        {
            ans.push_front(p);
        }
        p++;
    }

    while(!ans.empty())
    {
        cout << ans.front() << " ";
        ans.pop_front();
    }

    for(int i = 1; i < p; i++)
        cout << i << " ";
    return 0;
}