#include<iostream>
#include<deque>
using namespace std;
typedef long long LL;
LL N;
int tot=0;
deque <int> le,re;
inline void solve(LL n){
	if(n==1)
		return;
	solve(n>>1);
	le.push_front(++tot);
	re.push_front(tot);
	if(n&1){
		le.push_front(++tot);
		re.push_back(tot);
	}
}

int main(){
	cin>>N;
	solve(N+1);
	cout<<le.size()+re.size()<<endl;
	for(int i=0;i<le.size();i++)
		cout<<le[i]<<" ";
	for(int i=0;i<re.size();i++)
		cout<<re[i]<<" ";
	cout<<endl;
	return 0;
} 
### Prompt

Your task is to create 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<iostream>
#include<deque>
using namespace std;
typedef long long LL;
LL N;
int tot=0;
deque <int> le,re;
inline void solve(LL n){
	if(n==1)
		return;
	solve(n>>1);
	le.push_front(++tot);
	re.push_front(tot);
	if(n&1){
		le.push_front(++tot);
		re.push_back(tot);
	}
}

int main(){
	cin>>N;
	solve(N+1);
	cout<<le.size()+re.size()<<endl;
	for(int i=0;i<le.size();i++)
		cout<<le[i]<<" ";
	for(int i=0;i<re.size();i++)
		cout<<re[i]<<" ";
	cout<<endl;
	return 0;
} 
```