#include <iostream>

using namespace std;

long long N;

int a[200] = {0};
int b[200] = {0};

int numa = 1;
int numb = 1;

int big = 100;

int main()
{
	cin >> N;
	
	N ++;
	while(N > 1)
	{
		if(N % 2)
		{
			a[numa ++] = big --;
			N --;
		}
		else
		{
			b[numb ++] = big --;
			N /= 2;
		}	
	}
	int len = numa + numb - 2;
	len *= 2;
	cout << len << endl;
	for(int i = 103 - numa - numb; i <= 100; i ++)
		cout << i << " ";
	for(int i = 1; i < numa; i ++)
		cout << a[i] << " ";
	for(int i = numb - 1; i >= 1; i --)
		cout << b[i] << " ";
	return 0;	
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 <iostream>

using namespace std;

long long N;

int a[200] = {0};
int b[200] = {0};

int numa = 1;
int numb = 1;

int big = 100;

int main()
{
	cin >> N;
	
	N ++;
	while(N > 1)
	{
		if(N % 2)
		{
			a[numa ++] = big --;
			N --;
		}
		else
		{
			b[numb ++] = big --;
			N /= 2;
		}	
	}
	int len = numa + numb - 2;
	len *= 2;
	cout << len << endl;
	for(int i = 103 - numa - numb; i <= 100; i ++)
		cout << i << " ";
	for(int i = 1; i < numa; i ++)
		cout << a[i] << " ";
	for(int i = numb - 1; i >= 1; i --)
		cout << b[i] << " ";
	return 0;	
}
```