#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
#define MAXN 110
#define LL long long
int a[MAXN*2],l=MAXN,r=MAXN,ls=0;
void dfs(LL N)
{
	if(N==1)
	{
		a[l]=++ls;
		return;
	}
	if(N%2)
	{
		dfs(N/2);
		a[++r]=++ls;
	}
	else
	{
		dfs(N-1);
		a[--l]=++ls;
	}
}
LL n;
int main()
{
	scanf("%lld",&n);
	dfs(n);
	printf("%d\n",ls*2);
	for(int i=l;i<=r;i++)
		printf("%d ",a[i]);
	for(int i=1;i<=ls;i++)
		printf("%d ",i);
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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<algorithm>
#include<cmath>
using namespace std;
#define MAXN 110
#define LL long long
int a[MAXN*2],l=MAXN,r=MAXN,ls=0;
void dfs(LL N)
{
	if(N==1)
	{
		a[l]=++ls;
		return;
	}
	if(N%2)
	{
		dfs(N/2);
		a[++r]=++ls;
	}
	else
	{
		dfs(N-1);
		a[--l]=++ls;
	}
}
LL n;
int main()
{
	scanf("%lld",&n);
	dfs(n);
	printf("%d\n",ls*2);
	for(int i=l;i<=r;i++)
		printf("%d ",a[i]);
	for(int i=1;i<=ls;i++)
		printf("%d ",i);
}
```