#include <algorithm>
#include <iostream>
#include <string>
using namespace std;
typedef long long i64;
typedef int *pint;
const int MAXN = 100 + 20;
 
int a[MAXN], b[MAXN];
int p1, p2;
int M = 100;
i64 N;
 
int main()
{
	p1 = 0, p2 = 0;
	cin >> N, N++;
 
	while (N > 1)
	{
		if (N & 1)
			a[++p1] = M, M--, N--;
		else
			b[++p2] = M, M--, N >>= 1;
	}
 
	cout << ((p1 + p2) << 1) << endl;
 
	for (int i = 1; i <= p1; i++)
		cout << a[i] << " ";
	for (int i = p2; i >= 1; i--)
		cout << b[i] << " ";
	for (int i = 100 - p1 - p2 + 1; i <= 100; i++)
		cout << i << " ";
	cout << endl;
 
	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 <algorithm>
#include <iostream>
#include <string>
using namespace std;
typedef long long i64;
typedef int *pint;
const int MAXN = 100 + 20;
 
int a[MAXN], b[MAXN];
int p1, p2;
int M = 100;
i64 N;
 
int main()
{
	p1 = 0, p2 = 0;
	cin >> N, N++;
 
	while (N > 1)
	{
		if (N & 1)
			a[++p1] = M, M--, N--;
		else
			b[++p2] = M, M--, N >>= 1;
	}
 
	cout << ((p1 + p2) << 1) << endl;
 
	for (int i = 1; i <= p1; i++)
		cout << a[i] << " ";
	for (int i = p2; i >= 1; i--)
		cout << b[i] << " ";
	for (int i = 100 - p1 - p2 + 1; i <= 100; i++)
		cout << i << " ";
	cout << endl;
 
	return 0;
}
```