#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;

typedef long long s64;

s64 n;

int cur;
deque<int> result;

void solve(s64 n)
{
	if (n == 1ll)
		cur = 0;
	else
	{
		solve(n / 2);
		result.push_back(++cur);

		if (n & 1)
			result.push_front(++cur);
	}
}

int main()
{
	cin >> n;

	solve(n + 1);

	for (int i = 1; i <= cur; ++i)
		result.push_back(i);

	cout << result.size() << endl;
	for (int i : result)
		cout << i << ' ';
	cout << endl;

	return 0;
}
### 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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;

typedef long long s64;

s64 n;

int cur;
deque<int> result;

void solve(s64 n)
{
	if (n == 1ll)
		cur = 0;
	else
	{
		solve(n / 2);
		result.push_back(++cur);

		if (n & 1)
			result.push_front(++cur);
	}
}

int main()
{
	cin >> n;

	solve(n + 1);

	for (int i = 1; i <= cur; ++i)
		result.push_back(i);

	cout << result.size() << endl;
	for (int i : result)
		cout << i << ' ';
	cout << endl;

	return 0;
}
```