#include<stdio.h>
#include<vector>
#include<algorithm>
#include<deque>
using namespace std;
typedef long long ll;
int main()
{
	ll num;
	scanf("%lld", &num);
	num++;
	deque<int>v1, v2;
	vector<int>v;
	for (;;)
	{
		if (num == 1)break;
		v.push_back(num % 2);
		num /= 2;
	}
	int pt = 1;
	for (int i = int(v.size()) - 1; i >= 0; i--)
	{
		v1.push_back(pt);
		v2.push_back(pt);
		pt++;
		if (v[i])v1.push_back(pt), v2.push_front(pt), pt++;
	}
	printf("%d\n", v1.size() + v2.size());
	for (int i = 0; i < v1.size(); i++)printf("%d ", v1[i]);
	for (int i = 0; i < v2.size(); i++)printf("%d ", v2[i]);
	printf("\n");
}
### Prompt

Your challenge is to write 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<stdio.h>
#include<vector>
#include<algorithm>
#include<deque>
using namespace std;
typedef long long ll;
int main()
{
	ll num;
	scanf("%lld", &num);
	num++;
	deque<int>v1, v2;
	vector<int>v;
	for (;;)
	{
		if (num == 1)break;
		v.push_back(num % 2);
		num /= 2;
	}
	int pt = 1;
	for (int i = int(v.size()) - 1; i >= 0; i--)
	{
		v1.push_back(pt);
		v2.push_back(pt);
		pt++;
		if (v[i])v1.push_back(pt), v2.push_front(pt), pt++;
	}
	printf("%d\n", v1.size() + v2.size());
	for (int i = 0; i < v1.size(); i++)printf("%d ", v1[i]);
	for (int i = 0; i < v2.size(); i++)printf("%d ", v2[i]);
	printf("\n");
}
```