#include <iostream>
using namespace std;
int s[210];
int v[110];
int main()
{
	long long n;
	cin >> n;
	n ++;
	int ans = 0;
	int k = -1;
	while(n)

	{
		v[++ k] = n % 2;
		n /= 2;
	}
	for(int i = 1;i <= k;i ++)
		s[2 * i] = 1;
	for(int i = 0,j = 2 * k + 1;i < k;i ++,j -= 2)
		if(v[i])
			s[j] = 1;
	for(int i = 1;i <= 2 * k + 1;i ++)
		if(s[i])
			ans ++;
	cout << ans * 2 << endl;
	for(int i = 2 * k + 1;i > 0;i -= 2)
		if(s[i])
			cout << i << ' ';
	for(int i = 1;i <= k;i ++)
		cout << 2 * i << ' ';
	for(int i = 1;i <= 2 * k + 1;i ++)
		if(s[i])
			cout << i << ' ';
	cout << endl;
	return 0;
}
### Prompt

Please create a solution in Cpp 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 <iostream>
using namespace std;
int s[210];
int v[110];
int main()
{
	long long n;
	cin >> n;
	n ++;
	int ans = 0;
	int k = -1;
	while(n)

	{
		v[++ k] = n % 2;
		n /= 2;
	}
	for(int i = 1;i <= k;i ++)
		s[2 * i] = 1;
	for(int i = 0,j = 2 * k + 1;i < k;i ++,j -= 2)
		if(v[i])
			s[j] = 1;
	for(int i = 1;i <= 2 * k + 1;i ++)
		if(s[i])
			ans ++;
	cout << ans * 2 << endl;
	for(int i = 2 * k + 1;i > 0;i -= 2)
		if(s[i])
			cout << i << ' ';
	for(int i = 1;i <= k;i ++)
		cout << 2 * i << ' ';
	for(int i = 1;i <= 2 * k + 1;i ++)
		if(s[i])
			cout << i << ' ';
	cout << endl;
	return 0;
}
```