#include <bits/stdc++.h>

using namespace std;

typedef long long int64;

deque< int > latte, malta;
int num;

void dfs(int64 p)
{
  if(p == 1) return;
  if(p & 1) dfs(p - 1), latte.push_front(num), malta.push_back(num++);
  else dfs(p / 2), latte.push_back(num), malta.push_back(num++);
}

int main()
{
  int64 N;
  cin >> N;

  num = 1;
  dfs(++N);

  cout << latte.size() + malta.size() << endl;
  for(int i = 0; i < latte.size(); i++) {
    cout << latte[i] << " ";
  }
  for(int i = 0; i < malta.size(); i++) {
    cout << malta[i] << " ";
  }
  cout << endl;
}
### Prompt

Your task is to create a Cpp solution 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;

typedef long long int64;

deque< int > latte, malta;
int num;

void dfs(int64 p)
{
  if(p == 1) return;
  if(p & 1) dfs(p - 1), latte.push_front(num), malta.push_back(num++);
  else dfs(p / 2), latte.push_back(num), malta.push_back(num++);
}

int main()
{
  int64 N;
  cin >> N;

  num = 1;
  dfs(++N);

  cout << latte.size() + malta.size() << endl;
  for(int i = 0; i < latte.size(); i++) {
    cout << latte[i] << " ";
  }
  for(int i = 0; i < malta.size(); i++) {
    cout << malta[i] << " ";
  }
  cout << endl;
}
```