#include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;

int main(){
	ll n;cin>>n;n++;
	deque<int>f{1},r{1};
	int p=2;
	bool flag=false;
	for(int i=40;i>=0;i--){
		if(!flag){
			if(n>>i&1)flag=true;
			continue;
		}
		if(n>>i&1){//+1
			f.push_back(p);
			r.push_front(p);
			p++;
		}
		if(i){//*2
			f.push_back(p);
			r.push_back(p);
			p++;
		}
	}
	for(int i:r)f.push_back(i);
	cout<<f.size()<<endl;
	rep(i,f.size()){
		if(i)printf(" ");
		cout<<f[i];
	}
	cout<<endl;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;

int main(){
	ll n;cin>>n;n++;
	deque<int>f{1},r{1};
	int p=2;
	bool flag=false;
	for(int i=40;i>=0;i--){
		if(!flag){
			if(n>>i&1)flag=true;
			continue;
		}
		if(n>>i&1){//+1
			f.push_back(p);
			r.push_front(p);
			p++;
		}
		if(i){//*2
			f.push_back(p);
			r.push_back(p);
			p++;
		}
	}
	for(int i:r)f.push_back(i);
	cout<<f.size()<<endl;
	rep(i,f.size()){
		if(i)printf(" ");
		cout<<f[i];
	}
	cout<<endl;
}
```