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>

using namespace std;
typedef long long Int;

Int n;
Int pp = 1;
Int dp[210][210];

vector<Int> go(){
  vector<Int> res;
  Int k = 0;
  while((1LL << k) - 1 <= n)k++;
  Int p = 1;
  for(Int i = 0;i < k;i++)res.push_back(pp);pp++;
  n -= (1LL << (k-1)) - 1;
  for(Int i = k/2;i >= 0;i--){
    Int tmp = 0;
    for(Int j = 0;j <= k;j++){
      tmp += dp[i][j] * dp[k - i][j];
      if(tmp > n)break;
    }
    if(tmp > n)break;
    n -= tmp;
    res.insert(res.begin() + (p - 1), pp);
    res.insert(res.begin() + (i + p), pp);
    p++;
    pp++;
  }

  return res;
}

int main(){
  cin >> n;
  for(Int i = 0;i < 200;i++){
    for(Int j = 0;j < 200;j++){
      if(i < j)dp[i][j] = 0;
      else if(j == 0 || i == j)dp[i][j] = 1;
      else dp[i][j] = dp[i-1][j] + dp[i-1][j-1];
    }
  }

  vector<Int> res;
  
  while(n){
    cout << n << endl;
    vector<Int> tmp = go();
    res.insert(res.end(), tmp.begin(), tmp.end());
  }

  cout << res.size() << endl;
  for(Int i = 0;i < res.size();i++)cout << res[i] << " ";
  cout << endl;
  return 0;
}