#include <bits/stdc++.h>

using namespace std;

#define int long long

int n;
int id;
deque<int>S1,S2;
void Dfs(int x){
	if(x == 1)return;
	Dfs(x >> 1);
	id++;
	S1.push_front(id);
	S2.push_front(id);
	if(x & 1){
		id++;
		S1.push_back(id);
		S2.push_front(id);
	}
}
signed main(){
#ifdef Sirius
	freopen("in","r",stdin);
#endif 
	cin >> n;
	Dfs(n + 1);
	cout << S1.size() + S2.size() << endl;
	while(S1.size())cout << S1.front() << " ",S1.pop_front();
	while(S2.size())cout << S2.front() << " ",S2.pop_front();
	cout << endl;
	return 0;
}

### 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;

#define int long long

int n;
int id;
deque<int>S1,S2;
void Dfs(int x){
	if(x == 1)return;
	Dfs(x >> 1);
	id++;
	S1.push_front(id);
	S2.push_front(id);
	if(x & 1){
		id++;
		S1.push_back(id);
		S2.push_front(id);
	}
}
signed main(){
#ifdef Sirius
	freopen("in","r",stdin);
#endif 
	cin >> n;
	Dfs(n + 1);
	cout << S1.size() + S2.size() << endl;
	while(S1.size())cout << S1.front() << " ",S1.pop_front();
	while(S2.size())cout << S2.front() << " ",S2.pop_front();
	cout << endl;
	return 0;
}

```