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

typedef long long s64;
#define rep(i,l,r) for(int i=l;i<=r;++i)
vector<int>a;int mn=1,mx;
void solve(s64 n)
{
	if(!n)return ;
	if(n%2)
	{
		solve(n/2);
		a.push_back(++mx);
	}
	else
	{
		solve(n-1);
		a.push_back(--mn);
	}
}

int main()
{
	//freopen("1.in","r",stdin);
	s64 n;cin>>n;
	solve(n);
	printf("%d\n",a.size()*2);
	for(auto i:a)printf("%d ",i-mn+1);
	rep(i,1,a.size())printf("%d ",i);
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;

typedef long long s64;
#define rep(i,l,r) for(int i=l;i<=r;++i)
vector<int>a;int mn=1,mx;
void solve(s64 n)
{
	if(!n)return ;
	if(n%2)
	{
		solve(n/2);
		a.push_back(++mx);
	}
	else
	{
		solve(n-1);
		a.push_back(--mn);
	}
}

int main()
{
	//freopen("1.in","r",stdin);
	s64 n;cin>>n;
	solve(n);
	printf("%d\n",a.size()*2);
	for(auto i:a)printf("%d ",i-mn+1);
	rep(i,1,a.size())printf("%d ",i);
}
```