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 <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <unordered_map>
#include <set>
#include <climits>
#include <deque>
using namespace std;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    long long N;
    cin >> N;
    N++;
    int A, B;
    vector<int> ops;
    while(N != 1){
        if(N % 2 == 1){
            ops.push_back(0);
            N--;
        }else{
            N >>= 1;
            ops.push_back(1);
        }
    }
    reverse(ops.begin(), ops.end());

    int cnt = 0;
    deque<int> ans1, ans2;
    for(auto op : ops){
        if(op == 0){
            ans1.push_back(cnt);
            ans2.push_front(cnt);
        }else{
            ans1.push_back(cnt);
            ans2.push_back(cnt);
        }
        cnt++;
    }
    vector<int> ans;
    for(auto x : ans1) ans.push_back(x);
    for(auto x : ans2) ans.push_back(x);
    cout << ans.size() << endl;
    for(int i=0; i<ans.size()-1; i++)
        cout << ans[i] << " ";
    cout << ans.back() << endl;

    //int num = 0;
    //for(int i=1; i<(1<<ans.size()); i++){
    //    vector<int> v;
    //    for(int j=0; j<ans.size(); j++)
    //        if(i & (1<<j)) v.push_back(ans[j]);
    //    if(v.size() % 2 == 1) continue;
    //    bool flag = true;
    //    for(int j=0; j<v.size()/2; j++)
    //        if(v[j] != v[v.size()/2+j]){
    //            flag = false;
    //            break;
    //        }
    //    if(flag)
    //        num++;
    //}
    //cout << num << endl;
    return 0;
}