#include <bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef long double llf;
typedef pair<int, int> pi;
const int mod = 1e9 + 7;

vector<int> solve(lint x){
	if(x == 1) return vector<int>();
	if(x & 1){
		auto ret = solve(x-1);
		for(auto &i : ret) i++;
		ret.push_back(1);
		return ret;
	}
	auto ret = solve(x >> 1);
	ret.push_back(ret.size() + 1);
	return ret;
}

int main(){
	lint n;
	cin >> n;
	n++;
	vector<int> perm = solve(n);
	cout << 2 * perm.size() << endl;
	for(auto &i : perm) cout << i << " ";
	for(int i=1; i<=perm.size(); i++) cout << i << " ";
}

### 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 <bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef long double llf;
typedef pair<int, int> pi;
const int mod = 1e9 + 7;

vector<int> solve(lint x){
	if(x == 1) return vector<int>();
	if(x & 1){
		auto ret = solve(x-1);
		for(auto &i : ret) i++;
		ret.push_back(1);
		return ret;
	}
	auto ret = solve(x >> 1);
	ret.push_back(ret.size() + 1);
	return ret;
}

int main(){
	lint n;
	cin >> n;
	n++;
	vector<int> perm = solve(n);
	cout << 2 * perm.size() << endl;
	for(auto &i : perm) cout << i << " ";
	for(int i=1; i<=perm.size(); i++) cout << i << " ";
}

```