#include <bits/stdc++.h>
using namespace std;

long long N;
#define MAX_B 39
set<int> S;
int M, F;
deque<int> R;

int main() {
  cin >> N;
  for(int i = 0; i <= MAX_B; ++i) {
    if((N + 1) & (1LL << i)) S.insert(i);
  }
  int M = *S.rbegin(), F = M;
  for(int i = M - 1; i >= 0; --i) R.push_back(i);
  for(int i = M - 1; i >= 0; --i) {
    R.push_back(i);
    if(S.count(i)) {
      R.push_back(F); R.push_front(F); ++F;
    }
  }
  cout << R.size() << endl;
  for(int i = 0; i < (int)R.size(); ++i) {
    cout << (i == 0 ? "" : " ") << R[i] + 1;
  }
  cout << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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.

Examples

Input

7


Output

4
1 1 1 1


Input

299


Output

23
32 11 11 73 45 8 11 83 83 8 45 32 32 10 100 73 32 83 45 73 32 11 10

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

long long N;
#define MAX_B 39
set<int> S;
int M, F;
deque<int> R;

int main() {
  cin >> N;
  for(int i = 0; i <= MAX_B; ++i) {
    if((N + 1) & (1LL << i)) S.insert(i);
  }
  int M = *S.rbegin(), F = M;
  for(int i = M - 1; i >= 0; --i) R.push_back(i);
  for(int i = M - 1; i >= 0; --i) {
    R.push_back(i);
    if(S.count(i)) {
      R.push_back(F); R.push_front(F); ++F;
    }
  }
  cout << R.size() << endl;
  for(int i = 0; i < (int)R.size(); ++i) {
    cout << (i == 0 ? "" : " ") << R[i] + 1;
  }
  cout << endl;
  return 0;
}

```