#include<bits/stdc++.h>
const int MN=10000+2;
using namespace std;
typedef long long ll;
template<typename T>T IN(T&in){
	in=0;char c=getchar();int f=1;
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c))in=in*10+c-'0',c=getchar();
	in*=f;return in;
}
ll n;
int a[MN],L=100,R=L-1;
void input(){
	IN(n);n++;
	int p=40;
	while(!((1ll<<p)&n))p--;p--;
	while(p>=0){
		a[++R]=++a[0];
		if((1ll<<p)&n)a[--L]=++a[0];
		p--;
	}
	n=a[0];printf("%lld\n",n*2);
	for(int i=1;i<=n;++i)a[i]=a[i+L-1];
	for(int i=n+1;i<=n*2;++i)a[i]=i-n;
	for(int i=1;i<=n*2;++i)printf("%d ",a[i]);
}
int main(){
	
	input();
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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>
const int MN=10000+2;
using namespace std;
typedef long long ll;
template<typename T>T IN(T&in){
	in=0;char c=getchar();int f=1;
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c))in=in*10+c-'0',c=getchar();
	in*=f;return in;
}
ll n;
int a[MN],L=100,R=L-1;
void input(){
	IN(n);n++;
	int p=40;
	while(!((1ll<<p)&n))p--;p--;
	while(p>=0){
		a[++R]=++a[0];
		if((1ll<<p)&n)a[--L]=++a[0];
		p--;
	}
	n=a[0];printf("%lld\n",n*2);
	for(int i=1;i<=n;++i)a[i]=a[i+L-1];
	for(int i=n+1;i<=n*2;++i)a[i]=i-n;
	for(int i=1;i<=n*2;++i)printf("%d ",a[i]);
}
int main(){
	
	input();
	return 0;
}
```