#include "bits/stdc++.h"
using namespace std;
long long n;
vector < int > solve(long long n){
	if(n == 1){
		return vector < int > ();
	}
	if(n & 1){
		auto res = solve(n - 1);
		res.insert(res.begin() , res.size());
		return res;
	}
	else{
		auto res = solve(n >> 1);
		res.push_back(res.size());
		return res;
	}
}
int main(){
	cin >> n;
	auto ans = solve(n + 1);
	printf("%d\n" , int(ans.size()) * 2);
	for(int x : ans){
		cout << x + 1 << " ";
	}
	for(int i = 0 ; i < ans.size() ; ++i){
		cout << i + 1 << " ";
	}
	cout << endl;
}
### Prompt

In cpp, your task is to solve 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;
long long n;
vector < int > solve(long long n){
	if(n == 1){
		return vector < int > ();
	}
	if(n & 1){
		auto res = solve(n - 1);
		res.insert(res.begin() , res.size());
		return res;
	}
	else{
		auto res = solve(n >> 1);
		res.push_back(res.size());
		return res;
	}
}
int main(){
	cin >> n;
	auto ans = solve(n + 1);
	printf("%d\n" , int(ans.size()) * 2);
	for(int x : ans){
		cout << x + 1 << " ";
	}
	for(int i = 0 ; i < ans.size() ; ++i){
		cout << i + 1 << " ";
	}
	cout << endl;
}
```