#include <bits/stdc++.h>
using namespace std;
vector<int> solve(long long n, int tp){
	if(n == 1){
		vector<int> ans;
		return ans;
	}
	if(n & 1){
		vector<int> sol = solve(n - 1, tp + 1);
		sol.push_back(tp);
		return sol;
	}
	vector<int> sol = solve(n / 2, tp + 1);
	sol.insert(sol.begin(), tp);
	return sol;
}
int main(){
	long long n; cin >> n; n++;
	vector<int> sol = solve(n, 1);
	vector<int> cpy = sol;
	sort(cpy.begin(), cpy.end());
	cout << 2 * sol.size() << endl;
	for(int v : sol) cout << v << " ";
	for(int v : cpy) cout << v << " ";
	cout << endl;
	return 0;
}

### Prompt

Please create a solution in CPP to 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;
vector<int> solve(long long n, int tp){
	if(n == 1){
		vector<int> ans;
		return ans;
	}
	if(n & 1){
		vector<int> sol = solve(n - 1, tp + 1);
		sol.push_back(tp);
		return sol;
	}
	vector<int> sol = solve(n / 2, tp + 1);
	sol.insert(sol.begin(), tp);
	return sol;
}
int main(){
	long long n; cin >> n; n++;
	vector<int> sol = solve(n, 1);
	vector<int> cpy = sol;
	sort(cpy.begin(), cpy.end());
	cout << 2 * sol.size() << endl;
	for(int v : sol) cout << v << " ";
	for(int v : cpy) cout << v << " ";
	cout << endl;
	return 0;
}

```