#include<cstdio>
#include<cstring>
#include<algorithm>
#include<deque>
#define MAXN 200
using namespace std;
typedef long long LL;
LL N;
int len,digit[MAXN+5];
deque<int> que;
void Work(LL x)
{
	len=0;
	while(x)
	{
		digit[++len]=x%2;
		x/=2;
	}
}
int main()
{
	scanf("%lld",&N);
	N++;
	Work(N);
	int id=0;
	for(int i=len-1;i>=1;i--)
	{
		que.push_back(++id);
		if(digit[i]==1)
			que.push_front(++id);
	}
	printf("%d\n",id*2);
	for(int i=0;i<(int)que.size();i++)
		printf("%d ",que[i]);
	for(int i=1;i<=id;i++)
		if(i==1)
			printf("%d",i);
		else
			printf(" %d",i);
	printf("\n");
	return 0;
}
### Prompt

In Cpp, your task is to solve 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<algorithm>
#include<deque>
#define MAXN 200
using namespace std;
typedef long long LL;
LL N;
int len,digit[MAXN+5];
deque<int> que;
void Work(LL x)
{
	len=0;
	while(x)
	{
		digit[++len]=x%2;
		x/=2;
	}
}
int main()
{
	scanf("%lld",&N);
	N++;
	Work(N);
	int id=0;
	for(int i=len-1;i>=1;i--)
	{
		que.push_back(++id);
		if(digit[i]==1)
			que.push_front(++id);
	}
	printf("%d\n",id*2);
	for(int i=0;i<(int)que.size();i++)
		printf("%d ",que[i]);
	for(int i=1;i<=id;i++)
		if(i==1)
			printf("%d",i);
		else
			printf(" %d",i);
	printf("\n");
	return 0;
}
```