#include<bits/stdc++.h>
using namespace std;
#define LL long long

LL n;
vector<int> L,R;

int main()
{
#ifdef h10
	freopen("C.in","r",stdin);
	freopen("C.out","w",stdout);
#endif
	int i,pos=0;
	scanf("%lld",&n);
	n++;

	for (i=50;i>0;i--)
		if ((n>>i)&1) break;

	for (i--;i>=0;i--)
	{
		R.push_back(++pos);
		if ((n>>i)&1)
			L.push_back(++pos);
	}
	printf("%d\n",pos*2);
	for (i=1;i<=pos;i++)
		printf("%d ",i);
	for (i=(int)L.size()-1;i>=0;i--)
		printf("%d ",L[i]);
	for (i=0;i<(int)R.size();i++)
		printf("%d ",R[i]);
}

### 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<bits/stdc++.h>
using namespace std;
#define LL long long

LL n;
vector<int> L,R;

int main()
{
#ifdef h10
	freopen("C.in","r",stdin);
	freopen("C.out","w",stdout);
#endif
	int i,pos=0;
	scanf("%lld",&n);
	n++;

	for (i=50;i>0;i--)
		if ((n>>i)&1) break;

	for (i--;i>=0;i--)
	{
		R.push_back(++pos);
		if ((n>>i)&1)
			L.push_back(++pos);
	}
	printf("%d\n",pos*2);
	for (i=1;i<=pos;i++)
		printf("%d ",i);
	for (i=(int)L.size()-1;i>=0;i--)
		printf("%d ",L[i]);
	for (i=0;i<(int)R.size();i++)
		printf("%d ",R[i]);
}

```