#include<cstdio>
#include<cstdlib>
#define LL long long

int ans[210];
bool fec[110];

int main()
{
	LL n;
	scanf("%lld",&n);
	LL x=n;
	fec[0]=0;
	int mx=0,id,tot=0;
	for(int i=1;i<=40;i++) fec[i]=0;
	while(x)
	{
		if(x&1) fec[mx]=1;
		x>>=1;
		if(x) mx++;
	}
	id=mx;
	for(int i=0;i<=mx;i++)
	{
		if(i>0) ans[++tot]=i;
		if(fec[i]&&i!=mx) ans[++tot]=++id;
	}
	for(int i=1;i<=mx;i++) ans[++tot]=i;
	ans[++tot]=++id;ans[++tot]=id;
	for(int i=0;i<mx;i++) if(fec[i]) ans[++tot]=--id;
	printf("%d\n",tot);
	for(int i=1;i<=tot;i++) printf("%d ",ans[i]);
	return 0;
}
### 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<cstdio>
#include<cstdlib>
#define LL long long

int ans[210];
bool fec[110];

int main()
{
	LL n;
	scanf("%lld",&n);
	LL x=n;
	fec[0]=0;
	int mx=0,id,tot=0;
	for(int i=1;i<=40;i++) fec[i]=0;
	while(x)
	{
		if(x&1) fec[mx]=1;
		x>>=1;
		if(x) mx++;
	}
	id=mx;
	for(int i=0;i<=mx;i++)
	{
		if(i>0) ans[++tot]=i;
		if(fec[i]&&i!=mx) ans[++tot]=++id;
	}
	for(int i=1;i<=mx;i++) ans[++tot]=i;
	ans[++tot]=++id;ans[++tot]=id;
	for(int i=0;i<mx;i++) if(fec[i]) ans[++tot]=--id;
	printf("%d\n",tot);
	for(int i=1;i<=tot;i++) printf("%d ",ans[i]);
	return 0;
}
```