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

typedef long long LL;

LL N;

vector<int> Ans;
vector<int>::iterator it;

int main(){
	int i,j,now;
	scanf("%lld",&N);
	for(i=39;~N>>i&1;i--);
	for(j=1;j<=i;j++)
		Ans.push_back(j);
	for(j=1;j<=i;j++)
		Ans.push_back(j);
	Ans.push_back(i+1);
	Ans.push_back(i+1);
	now=i+1;
	for(i--;~i;i--)
		if(N>>i&1){
			it=find(Ans.begin(),Ans.end(),Ans[i]);
			Ans.insert(it,now+1);
			it=find(Ans.begin(),Ans.end(),Ans[now<<1]);
			Ans.insert(it,now+1);
			now++;
		}
	printf("%d\n",Ans.size());
	for(i=0;i<Ans.size();i++)
		printf("%d ",Ans[i]);
	return 0;
}
### 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>
using namespace std;

typedef long long LL;

LL N;

vector<int> Ans;
vector<int>::iterator it;

int main(){
	int i,j,now;
	scanf("%lld",&N);
	for(i=39;~N>>i&1;i--);
	for(j=1;j<=i;j++)
		Ans.push_back(j);
	for(j=1;j<=i;j++)
		Ans.push_back(j);
	Ans.push_back(i+1);
	Ans.push_back(i+1);
	now=i+1;
	for(i--;~i;i--)
		if(N>>i&1){
			it=find(Ans.begin(),Ans.end(),Ans[i]);
			Ans.insert(it,now+1);
			it=find(Ans.begin(),Ans.end(),Ans[now<<1]);
			Ans.insert(it,now+1);
			now++;
		}
	printf("%d\n",Ans.size());
	for(i=0;i<Ans.size();i++)
		printf("%d ",Ans[i]);
	return 0;
}
```