#include<cstdio>
#include<deque>
#include<vector>
#include<algorithm>
using namespace std;

typedef long long ll;

int main(){
	ll N; scanf("%lld", &N);

	vector<char> v;
	while (N){
		if (N % 2){
			N = (N - 1) / 2;
			v.push_back('R');
		}
		else{
			N--;
			v.push_back('L');
		}
	}

	reverse(v.begin(), v.end());

	deque<int> dq;
	for (int i = 0; i < v.size(); i++){
		char q = v[i];
		if (q == 'L') dq.push_front(i + 1);
		else dq.push_back(i + 1);
	}

	printf("%d\n", dq.size() * 2);
	for (int i = 1; i <= dq.size(); i++) printf("%d ", i);
	for (int x : dq){
		printf("%d ", x);
	}
	printf("\n");
}
### Prompt

Construct a cpp code solution to the problem outlined:

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<cstdio>
#include<deque>
#include<vector>
#include<algorithm>
using namespace std;

typedef long long ll;

int main(){
	ll N; scanf("%lld", &N);

	vector<char> v;
	while (N){
		if (N % 2){
			N = (N - 1) / 2;
			v.push_back('R');
		}
		else{
			N--;
			v.push_back('L');
		}
	}

	reverse(v.begin(), v.end());

	deque<int> dq;
	for (int i = 0; i < v.size(); i++){
		char q = v[i];
		if (q == 'L') dq.push_front(i + 1);
		else dq.push_back(i + 1);
	}

	printf("%d\n", dq.size() * 2);
	for (int i = 1; i <= dq.size(); i++) printf("%d ", i);
	for (int x : dq){
		printf("%d ", x);
	}
	printf("\n");
}
```