#include<iostream>
#include<cstdio>
#include<set>
#define mk make_pair
using namespace std;
long long  n;
set<pair<int,int> > s;
int main()
{
	scanf("%lld",&n);
	int ma;
	for(ma=1;(1LL<<ma)<n;ma++);ma--;
	for(int i=1;i<=ma;i++)
	s.insert(mk(i,i));
	for(int i=1;i<=ma;i++)
	s.insert(mk(i+ma,i));
	s.insert(mk(ma+ma+1,ma+1));
	s.insert(mk(ma+ma+2,ma+1));
	int sz=ma+ma+2,nm=ma+1;
	for(int i=ma-1;~i;i--)
	if(n&(1LL<<i))
	{
		nm++;
		s.insert(mk(i,nm));
		s.insert(mk(sz,nm));
	}
	printf("%d\n",s.size());
	for(set<pair<int,int> >::iterator it=s.begin();it!=s.end();it++)
	printf("%d ",(*it).second);
	return 0;
}
	
### 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<iostream>
#include<cstdio>
#include<set>
#define mk make_pair
using namespace std;
long long  n;
set<pair<int,int> > s;
int main()
{
	scanf("%lld",&n);
	int ma;
	for(ma=1;(1LL<<ma)<n;ma++);ma--;
	for(int i=1;i<=ma;i++)
	s.insert(mk(i,i));
	for(int i=1;i<=ma;i++)
	s.insert(mk(i+ma,i));
	s.insert(mk(ma+ma+1,ma+1));
	s.insert(mk(ma+ma+2,ma+1));
	int sz=ma+ma+2,nm=ma+1;
	for(int i=ma-1;~i;i--)
	if(n&(1LL<<i))
	{
		nm++;
		s.insert(mk(i,nm));
		s.insert(mk(sz,nm));
	}
	printf("%d\n",s.size());
	for(set<pair<int,int> >::iterator it=s.begin();it!=s.end();it++)
	printf("%d ",(*it).second);
	return 0;
}
	
```