#include <iostream>
#include <vector>
using namespace std;

long N, n;
vector<int> p;

int main(void) {
  cin >> N;

  for(n = 50; n > 0; n--) if(N >= (1L << n) - 1) break;
  N -= (1L << n) - 1;

  for(int i = 1; i <= n; i++) p.push_back(i);
  for(int i = 1; i <= n; i++) p.push_back(i);

  for(int i = n - 1; i >= 0; i--)
    if(N >= (1L << i)) {
      N -= (1L << i);
      n++;
      p.insert(p.begin() + i, n);
      p.push_back(n);
    }

  cout << p.size() << endl;
  for(int i = 0; i < p.size(); i++) cout << p[i] << " ";
  cout << endl;

  return 0;
}

### Prompt

Please formulate 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 <iostream>
#include <vector>
using namespace std;

long N, n;
vector<int> p;

int main(void) {
  cin >> N;

  for(n = 50; n > 0; n--) if(N >= (1L << n) - 1) break;
  N -= (1L << n) - 1;

  for(int i = 1; i <= n; i++) p.push_back(i);
  for(int i = 1; i <= n; i++) p.push_back(i);

  for(int i = n - 1; i >= 0; i--)
    if(N >= (1L << i)) {
      N -= (1L << i);
      n++;
      p.insert(p.begin() + i, n);
      p.push_back(n);
    }

  cout << p.size() << endl;
  for(int i = 0; i < p.size(); i++) cout << p[i] << " ";
  cout << endl;

  return 0;
}

```