#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef deque<ll> V;
V dfs(ll n){
  if(n==1)return V();
  V res=dfs(n/2),t;
  res.push_back(res.size()+1);
  if(n%2)res.push_front(res.size()+1);
  return res;
}

int main(){
  ll n;
  cin>>n;
  V res=dfs(n+1);
  cout<<res.size()*2<<endl;
  for(int i=1;i<=res.size();i++){
    cout<<i;
    if(i!=res.size())cout<<" ";
  }
  while(!res.empty())
    cout<<" "<<res.front(),res.pop_front();
  
  cout<<endl;
  return 0;
}
### 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 ll;
typedef deque<ll> V;
V dfs(ll n){
  if(n==1)return V();
  V res=dfs(n/2),t;
  res.push_back(res.size()+1);
  if(n%2)res.push_front(res.size()+1);
  return res;
}

int main(){
  ll n;
  cin>>n;
  V res=dfs(n+1);
  cout<<res.size()*2<<endl;
  for(int i=1;i<=res.size();i++){
    cout<<i;
    if(i!=res.size())cout<<" ";
  }
  while(!res.empty())
    cout<<" "<<res.front(),res.pop_front();
  
  cout<<endl;
  return 0;
}
```