#include<bits/stdc++.h>
#define LL long long
#define db double

using namespace std;
const int N=400+10;
LL rd()
{
    LL x=0,w=1;char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch^48);ch=getchar();}
    return x*w;
}
int p[N],hd,tl,nt;

int main()
{
	LL rs=rd()+1;
	hd=201,tl=200,nt=0;
	int w=50;
	while(!(rs>>w&1)) --w;
	for(int i=w-1;~i;--i)
	{
		++nt,p[++tl]=nt;
		if(rs>>i&1) ++nt,p[--hd]=nt;
	}
	printf("%d\n",(nt<<1));
	for(int i=1;i<=nt;++i) printf("%d ",i);
	for(int i=hd;i<=tl;++i) printf("%d ",p[i]);
	return 0;
}

### Prompt

Your task is to create 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 LL long long
#define db double

using namespace std;
const int N=400+10;
LL rd()
{
    LL x=0,w=1;char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch^48);ch=getchar();}
    return x*w;
}
int p[N],hd,tl,nt;

int main()
{
	LL rs=rd()+1;
	hd=201,tl=200,nt=0;
	int w=50;
	while(!(rs>>w&1)) --w;
	for(int i=w-1;~i;--i)
	{
		++nt,p[++tl]=nt;
		if(rs>>i&1) ++nt,p[--hd]=nt;
	}
	printf("%d\n",(nt<<1));
	for(int i=1;i<=nt;++i) printf("%d ",i);
	for(int i=hd;i<=tl;++i) printf("%d ",p[i]);
	return 0;
}

```