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

typedef long long LL;

LL n;
deque<int> ans;

int main() {
	cin >> n;
	++n;
	int cur = 0;
	for (int i = 40; i >= 0; --i) {
		if (n>>i > 1) {
			ans.push_back(++cur);
			if (n>>i&1) {
				ans.push_front(++cur);
			}
		}
	}
	cout << 2*cur << endl;
	for (deque<int>::iterator it = ans.begin(); it != ans.end(); ++it) {
		cout << *it << ' ';
	}
	for (int i = 1; i <= cur; ++i) {
		cout << i << " \n"[i == cur];
	}
	return 0;
}
### Prompt

Generate 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;

typedef long long LL;

LL n;
deque<int> ans;

int main() {
	cin >> n;
	++n;
	int cur = 0;
	for (int i = 40; i >= 0; --i) {
		if (n>>i > 1) {
			ans.push_back(++cur);
			if (n>>i&1) {
				ans.push_front(++cur);
			}
		}
	}
	cout << 2*cur << endl;
	for (deque<int>::iterator it = ans.begin(); it != ans.end(); ++it) {
		cout << *it << ' ';
	}
	for (int i = 1; i <= cur; ++i) {
		cout << i << " \n"[i == cur];
	}
	return 0;
}
```