#include<iostream>
#include<vector>
using namespace std;

vector<int> ans;
int cnt;
void construct(long long n,int &k)
{
	if(n==1)
	{
		ans.push_back(1);
		k=1;
		return;
	}
	if(n%2)
	{
		construct((n-1)/2,k);
		k++;
		ans.push_back(k);
	}
	else 
	{
		construct(n-1,k);
		k++;
		ans.insert(ans.begin(),k);
	}
}

int main()
{	
	long long n;
	cin>>n;
	construct(n,cnt);
	for(int i=1;i<=cnt;i++)
		ans.push_back(i);
	cout<<ans.size()<<endl;
	for(int i=0;i<ans.size();i++)
	{
		cout<<ans[i]<<' ';
	}
	return 0;
}
### 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<iostream>
#include<vector>
using namespace std;

vector<int> ans;
int cnt;
void construct(long long n,int &k)
{
	if(n==1)
	{
		ans.push_back(1);
		k=1;
		return;
	}
	if(n%2)
	{
		construct((n-1)/2,k);
		k++;
		ans.push_back(k);
	}
	else 
	{
		construct(n-1,k);
		k++;
		ans.insert(ans.begin(),k);
	}
}

int main()
{	
	long long n;
	cin>>n;
	construct(n,cnt);
	for(int i=1;i<=cnt;i++)
		ans.push_back(i);
	cout<<ans.size()<<endl;
	for(int i=0;i<ans.size();i++)
	{
		cout<<ans[i]<<' ';
	}
	return 0;
}
```