# include <bits/stdc++.h>

using namespace std;

long long N;

vector<int> ans;

int m;
deque<int> q;

void get(long long N) {
	if(N == 0) return;
	if(N & 1) {
		get(N >> 1);
		q.push_back(q.size() + 1);
	} else {
		get(N - 1);
		q.push_front(q.size() + 1);
	}
}

int main() {
	cin >> N;
	get(N);
	int n = q.size();
	ans.resize(n * 2);
	for(int i = 0; i < n; ++i) {
		ans[i] = *q.begin();
		q.pop_front();
		ans[i + n] = i + 1;
	}
	printf("%d\n", n * 2);
	for(int i = 0; i < n * 2; ++i) {
		printf("%d%c", ans[i], " \n"[i==n*2]);
	}
	return 0;
}

### Prompt

Your task is to create a Cpp solution 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;

long long N;

vector<int> ans;

int m;
deque<int> q;

void get(long long N) {
	if(N == 0) return;
	if(N & 1) {
		get(N >> 1);
		q.push_back(q.size() + 1);
	} else {
		get(N - 1);
		q.push_front(q.size() + 1);
	}
}

int main() {
	cin >> N;
	get(N);
	int n = q.size();
	ans.resize(n * 2);
	for(int i = 0; i < n; ++i) {
		ans[i] = *q.begin();
		q.pop_front();
		ans[i + n] = i + 1;
	}
	printf("%d\n", n * 2);
	for(int i = 0; i < n * 2; ++i) {
		printf("%d%c", ans[i], " \n"[i==n*2]);
	}
	return 0;
}

```