#include <cstdio>
#include <iostream>
#include <algorithm>
#include <queue>

#define For(i, j, k) for(int i = j; i <= k; i++)

using namespace std;

deque<int> A, B;
int c;

void work(long long n){
	if(n == 1) return;
	if(n > 1){
		work(n >> 1);
		A.push_back(++c), B.push_back(c);
	}
	if(n & 1) A.push_back(++c), B.push_front(c);
}

long long n;

int main(){
	cin >> n;
	work(n + 1);
	int x = A.size() + B.size();
	printf("%d\n", x);
	for(int v : A) printf("%d ", v);
	for(int v : B) printf("%d ", v);
	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 <cstdio>
#include <iostream>
#include <algorithm>
#include <queue>

#define For(i, j, k) for(int i = j; i <= k; i++)

using namespace std;

deque<int> A, B;
int c;

void work(long long n){
	if(n == 1) return;
	if(n > 1){
		work(n >> 1);
		A.push_back(++c), B.push_back(c);
	}
	if(n & 1) A.push_back(++c), B.push_front(c);
}

long long n;

int main(){
	cin >> n;
	work(n + 1);
	int x = A.size() + B.size();
	printf("%d\n", x);
	for(int v : A) printf("%d ", v);
	for(int v : B) printf("%d ", v);
	return 0;
}
```