#include <bits/stdc++.h>

using namespace std;

int main() {

	long long n;
	scanf("%lld", &n);


	vector <int> a;

	while(n) {
		if(n%2==1) { //2*n+1
			a.push_back(0);
			n/=2;
		} else { // n+1
			a.push_back(1);
			n--;
		}
	}

	reverse(a.begin(), a.end());
	int cnt = 1;

	deque <int> ans;

	for (int i = 0; i < a.size(); ++i) {
		if(a[i] == 0) {
			ans.push_back(cnt++);
		} else {
			ans.push_front(cnt++);
		}
	}

	printf("%lu\n", 2*ans.size());
	for (int x: ans) {
		printf("%d ", x);
	}
	sort(ans.begin(), ans.end());
	for (int x: ans) {
		printf("%d ", x);
	}

}
### Prompt

In Cpp, your task is to solve 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;

int main() {

	long long n;
	scanf("%lld", &n);


	vector <int> a;

	while(n) {
		if(n%2==1) { //2*n+1
			a.push_back(0);
			n/=2;
		} else { // n+1
			a.push_back(1);
			n--;
		}
	}

	reverse(a.begin(), a.end());
	int cnt = 1;

	deque <int> ans;

	for (int i = 0; i < a.size(); ++i) {
		if(a[i] == 0) {
			ans.push_back(cnt++);
		} else {
			ans.push_front(cnt++);
		}
	}

	printf("%lu\n", 2*ans.size());
	for (int x: ans) {
		printf("%d ", x);
	}
	sort(ans.begin(), ans.end());
	for (int x: ans) {
		printf("%d ", x);
	}

}
```