#include <iostream>
#include <vector>
#define llint long long

using namespace std;

llint n;
vector<int> vec, vec2;

int main(void)
{
	cin >> n;
	n++;
	
	int keta = 0;
	for(llint t = n; t; t/=2) keta++;
	
	int id = 1;
	for(int i = keta-2; i >= 0; i--){
		vec.push_back(id), vec2.push_back(id++);
		if(n & (1LL<<i)) vec.insert(vec.begin(), id), vec2.push_back(id++);
	}
	cout << vec.size()*2 << endl;
	for(int i = 0; i < vec.size(); i++) cout << vec[i] << " ";
	for(int i = 0; i < vec2.size(); i++) cout << vec2[i] << " ";
	
	return 0;
}
### 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 <iostream>
#include <vector>
#define llint long long

using namespace std;

llint n;
vector<int> vec, vec2;

int main(void)
{
	cin >> n;
	n++;
	
	int keta = 0;
	for(llint t = n; t; t/=2) keta++;
	
	int id = 1;
	for(int i = keta-2; i >= 0; i--){
		vec.push_back(id), vec2.push_back(id++);
		if(n & (1LL<<i)) vec.insert(vec.begin(), id), vec2.push_back(id++);
	}
	cout << vec.size()*2 << endl;
	for(int i = 0; i < vec.size(); i++) cout << vec[i] << " ";
	for(int i = 0; i < vec2.size(); i++) cout << vec2[i] << " ";
	
	return 0;
}
```