#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int i,j,k,prt[1005],cnt,num;
long long n;
int main()
{
	scanf("%lld",&n);
	++n;
	for(k=60;!(n>>k&1);--k);
	--k;
	while(k>=0)
	{
		++num;
		for(i=cnt;i>cnt/2;--i)
			prt[i+1]=prt[i];
		prt[cnt/2+1]=num;
		++cnt;
		for(i=cnt;i>=1;--i)
			prt[i+1]=prt[i];
		++cnt;
		prt[1]=num;
		if(n>>k&1)
		{
			for(i=cnt;i>=1;--i)
				prt[i+1]=prt[i];
			++cnt;
			prt[1]=prt[++cnt]=++num;
		}
		--k;
	}
	printf("%d\n",cnt);
	for(i=1;i<=cnt;++i)
		printf("%d ",prt[i]);
	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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int i,j,k,prt[1005],cnt,num;
long long n;
int main()
{
	scanf("%lld",&n);
	++n;
	for(k=60;!(n>>k&1);--k);
	--k;
	while(k>=0)
	{
		++num;
		for(i=cnt;i>cnt/2;--i)
			prt[i+1]=prt[i];
		prt[cnt/2+1]=num;
		++cnt;
		for(i=cnt;i>=1;--i)
			prt[i+1]=prt[i];
		++cnt;
		prt[1]=num;
		if(n>>k&1)
		{
			for(i=cnt;i>=1;--i)
				prt[i+1]=prt[i];
			++cnt;
			prt[1]=prt[++cnt]=++num;
		}
		--k;
	}
	printf("%d\n",cnt);
	for(i=1;i<=cnt;++i)
		printf("%d ",prt[i]);
	return 0;
}

```