#include<bits/stdc++.h>
#define int long long
using namespace std;
vector<int> ans1,ans2;
void a(int n,int k){
  if(n==1)return;
  if(n%2){
    a(n-1,k+1);
    ans1.push_back(k);
  }else{
    a(n/2,k+1);
    ans2.push_back(k);
  }
}
signed main(){
  int N;
  cin>>N;
  a(N+1,1);
  reverse(ans1.begin(),ans1.end());
  for(int i:ans2)ans1.push_back(i);
  int M=ans1.size();
  cout<<M*2<<endl;
  for(int i=0;i<M;i++)cout<<ans1[i]<<' ';
  for(int i=M;i>=1;i--)cout<<i<<(i==1?'\n':' ');
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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>
#define int long long
using namespace std;
vector<int> ans1,ans2;
void a(int n,int k){
  if(n==1)return;
  if(n%2){
    a(n-1,k+1);
    ans1.push_back(k);
  }else{
    a(n/2,k+1);
    ans2.push_back(k);
  }
}
signed main(){
  int N;
  cin>>N;
  a(N+1,1);
  reverse(ans1.begin(),ans1.end());
  for(int i:ans2)ans1.push_back(i);
  int M=ans1.size();
  cout<<M*2<<endl;
  for(int i=0;i<M;i++)cout<<ans1[i]<<' ';
  for(int i=M;i>=1;i--)cout<<i<<(i==1?'\n':' ');
}
```