#include<bits/stdc++.h>
using namespace std;

vector<int> solve(long long int x){
	if(x==1)	return vector<int>();
	if(x&1){
		auto res = solve(x-1);
		res.insert(res.begin(),res.size()+1);
		return res;
	}
	else{
		auto res = solve(x>>1);
		res.emplace_back(res.size()+1);
		return res;
	}
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	long long int n;
	cin>>n;
	n++;
	auto res = solve(n);
	cout<<res.size()*2<<endl;
	for(int it:res)
		cout<<it<<' ';
	for(int i=1;i<=res.size();i++)
		cout<<i<<' ';
	cout<<endl;
}

### Prompt

Create a solution in cpp for 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;

vector<int> solve(long long int x){
	if(x==1)	return vector<int>();
	if(x&1){
		auto res = solve(x-1);
		res.insert(res.begin(),res.size()+1);
		return res;
	}
	else{
		auto res = solve(x>>1);
		res.emplace_back(res.size()+1);
		return res;
	}
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	long long int n;
	cin>>n;
	n++;
	auto res = solve(n);
	cout<<res.size()*2<<endl;
	for(int it:res)
		cout<<it<<' ';
	for(int i=1;i<=res.size();i++)
		cout<<i<<' ';
	cout<<endl;
}

```