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 <algorithm>
#include <fstream>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <set>
#include <stdio.h>
#include <string>
#include <utility>
#include <vector>

using namespace std;

int main() {
  long long n;
  cin >> n;
  vector<int> p;
  for (int i = 0; (1LL << (i + 1)) <= n + 1; i++)
    p.push_back(i);
  long long plen = p.size();
  long long m = n - ((1LL << plen) - 1);
  set<int> used;
  for (int k = plen - 1; k >= 0; k--) {
    if ((m >> k) & 1) {
      p.insert(p.begin() + k, 99 - k);
      used.insert(99 - k);
    }
  }
  vector<int> q;
  cout << p.size() * 2 << endl;
  for (int i = 0; i < p.size(); i++) {
    q.push_back(p[i] + 1);
  }
  for (int i = 0; i < plen; i++)
    q.push_back(i + 1);
  for (auto &i : used)
    q.push_back(i + 1);
  cout << q[0];
  for (int i = 1; i < q.size(); i++)
    cout << " " << q[i];
  cout << endl;
  return 0;
}