#include <cstdio>
#include <deque>

using namespace std;

deque<int> q1, q2;
int ans = 0;

void fun(long long n){
	if(n == 1)
		return;
	fun(n >> 1);
	ans++;
	q1.push_front(ans);
	q2.push_front(ans);
	if(n & 1){
		ans++;
		q1.push_front(ans);
		q2.push_back(ans);
	}
	return;
}

int main(){
	long long n;
	scanf("%lld", &n);
	fun(n + 1);
	printf("%d\n", q1.size() + q2.size());
	while(!q1.empty()){
		printf("%d ", q1.front());
		q1.pop_front();
	}
	while(!q2.empty()){
		printf("%d ", q2.front());
		q2.pop_front();
	}
	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 <cstdio>
#include <deque>

using namespace std;

deque<int> q1, q2;
int ans = 0;

void fun(long long n){
	if(n == 1)
		return;
	fun(n >> 1);
	ans++;
	q1.push_front(ans);
	q2.push_front(ans);
	if(n & 1){
		ans++;
		q1.push_front(ans);
		q2.push_back(ans);
	}
	return;
}

int main(){
	long long n;
	scanf("%lld", &n);
	fun(n + 1);
	printf("%d\n", q1.size() + q2.size());
	while(!q1.empty()){
		printf("%d ", q1.front());
		q1.pop_front();
	}
	while(!q2.empty()){
		printf("%d ", q2.front());
		q2.pop_front();
	}
	return 0;
} 
```