#include<bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
  int n;
  cin>>n;
  int p=0,q=1;
  while(q*2-1<=n) q*=2,p++;
  vector<int> ans;
  for(int i=0;i<p;i++) ans.push_back(i+1);
  q=n-(q-1);
  int x=1;
  for(int i=p-1;i>=0;i--){
    if((q>>i)&1){
      ans.insert(ans.begin()+i,p+x);
      x++;
    }
  }
  q=ans.size();
  for(int i=0;i<q;i++) ans.push_back(i+1);
  cout<<ans.size()<<endl;
  for(int i=0;i<(int)ans.size();i++)
    cout<<ans[i]<<" \n"[i==(int)ans.size()-1];
  return 0;
}

### Prompt

Generate 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;
#define int long long
signed main(){
  int n;
  cin>>n;
  int p=0,q=1;
  while(q*2-1<=n) q*=2,p++;
  vector<int> ans;
  for(int i=0;i<p;i++) ans.push_back(i+1);
  q=n-(q-1);
  int x=1;
  for(int i=p-1;i>=0;i--){
    if((q>>i)&1){
      ans.insert(ans.begin()+i,p+x);
      x++;
    }
  }
  q=ans.size();
  for(int i=0;i<q;i++) ans.push_back(i+1);
  cout<<ans.size()<<endl;
  for(int i=0;i<(int)ans.size();i++)
    cout<<ans[i]<<" \n"[i==(int)ans.size()-1];
  return 0;
}

```