#include <bits/stdc++.h>
using namespace std;
typedef long long LL;


int main(){
	LL n;
	cin >> n;
	vector<int> q;
	for(int i = 1; i <= 100; i++){
		q.push_back(i);
	}
	vector<int> r;
	for(LL i = 0; i < 100; i++){
		if(n == 0) break;
		LL g = 1;
		LL num = 0;
		for(int f = 0; f < i; f++){
			if(g * 2 > n) break;
			g *= 2;
			num ++;
		}
		n -= g;
		for(int f = 0; f < r.size(); f++){
			if(r[f] >= num + 1) r[f]++;
		}
		r.push_back(num+1);
	}
	for(int i = 0; i < r.size(); i++){
		q.push_back(r[i]);
	}
	cout << q.size() << endl;
	for(int i = 0; i < q.size(); i++){
		cout << q[i] << " ";
	}
	cout << endl;
}
### Prompt

Develop a solution in cpp to the problem described below:

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;
typedef long long LL;


int main(){
	LL n;
	cin >> n;
	vector<int> q;
	for(int i = 1; i <= 100; i++){
		q.push_back(i);
	}
	vector<int> r;
	for(LL i = 0; i < 100; i++){
		if(n == 0) break;
		LL g = 1;
		LL num = 0;
		for(int f = 0; f < i; f++){
			if(g * 2 > n) break;
			g *= 2;
			num ++;
		}
		n -= g;
		for(int f = 0; f < r.size(); f++){
			if(r[f] >= num + 1) r[f]++;
		}
		r.push_back(num+1);
	}
	for(int i = 0; i < r.size(); i++){
		q.push_back(r[i]);
	}
	cout << q.size() << endl;
	for(int i = 0; i < q.size(); i++){
		cout << q[i] << " ";
	}
	cout << endl;
}
```