#include<bits/stdc++.h>
#define L long long
using namespace std;
int m,x[210],p;
L n;
inline void dfs(L n)
{
    if(n==1)
      return;
    if(n&1)
      {
       int u=++p;
       dfs(n-1);
       x[++m]=u;
      }
    else
      {
       x[++m]=++p;
       dfs(n/2);
      }
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i;
	cin>>n;
	n++;
	dfs(n);
	printf("%d\n",2*m);
	for(i=1;i<=m;i++)
	  printf("%d ",i);
	for(i=1;i<=m;i++)
	  printf("%d ",x[i]);
	return 0;
}

### Prompt

Generate 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<bits/stdc++.h>
#define L long long
using namespace std;
int m,x[210],p;
L n;
inline void dfs(L n)
{
    if(n==1)
      return;
    if(n&1)
      {
       int u=++p;
       dfs(n-1);
       x[++m]=u;
      }
    else
      {
       x[++m]=++p;
       dfs(n/2);
      }
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i;
	cin>>n;
	n++;
	dfs(n);
	printf("%d\n",2*m);
	for(i=1;i<=m;i++)
	  printf("%d ",i);
	for(i=1;i<=m;i++)
	  printf("%d ",x[i]);
	return 0;
}

```