#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long LL;

int a[305];
LL bin[55],n;

int main()
{
	scanf("%lld",&n);
	bin[0]=1;
	for (int i=1;i<=50;i++) bin[i]=bin[i-1]*2;
	int m=50;
	while (bin[m]>n) m--;
	int tot=0,p=m+1;
	for (int i=1;i<=m;i++) a[++tot]=i;
	for (int i=1;i<=m;i++) a[++tot]=i;
	a[++tot]=m+1;a[++tot]=m+1;
	for (int i=m-1;i>=0;i--)
		if (n&bin[i])
		{
			tot++;
			for (int j=tot;j>i+1;j--) a[j]=a[j-1];
			p++;a[i+1]=a[++tot]=p;
		}
	printf("%d\n",tot);
	for (int i=1;i<=tot;i++) printf("%d ",a[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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long LL;

int a[305];
LL bin[55],n;

int main()
{
	scanf("%lld",&n);
	bin[0]=1;
	for (int i=1;i<=50;i++) bin[i]=bin[i-1]*2;
	int m=50;
	while (bin[m]>n) m--;
	int tot=0,p=m+1;
	for (int i=1;i<=m;i++) a[++tot]=i;
	for (int i=1;i<=m;i++) a[++tot]=i;
	a[++tot]=m+1;a[++tot]=m+1;
	for (int i=m-1;i>=0;i--)
		if (n&bin[i])
		{
			tot++;
			for (int j=tot;j>i+1;j--) a[j]=a[j-1];
			p++;a[i+1]=a[++tot]=p;
		}
	printf("%d\n",tot);
	for (int i=1;i<=tot;i++) printf("%d ",a[i]);
	return 0;
}
```