#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=100005;
LL read(){
	LL x=0;
	char ch=getchar();
	while (!isdigit(ch))
		ch=getchar();
	while (isdigit(ch))
		x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return x;
}
LL n;
int a[N],tot;
int main(){
	n=read()+1;
	tot=100;
	for (int i=1;i<=tot;i++)
		a[i]=i;
	int t=0;
	while (n>>(t+1))
		t++;
	for (int i=1;i<=t;i++)
		a[++tot]=i;
	int k=t;
	for (int x=t-1;x>=0;x--)
		if (n>>x&1){
			tot++;
			for (int i=tot;i>101+x;i--)
				a[i]=a[i-1];
			a[101+x]=++k;
		}
	cout << tot << endl;
	for (int i=1;i<=tot;i++)
		cout << a[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 <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=100005;
LL read(){
	LL x=0;
	char ch=getchar();
	while (!isdigit(ch))
		ch=getchar();
	while (isdigit(ch))
		x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return x;
}
LL n;
int a[N],tot;
int main(){
	n=read()+1;
	tot=100;
	for (int i=1;i<=tot;i++)
		a[i]=i;
	int t=0;
	while (n>>(t+1))
		t++;
	for (int i=1;i<=t;i++)
		a[++tot]=i;
	int k=t;
	for (int x=t-1;x>=0;x--)
		if (n>>x&1){
			tot++;
			for (int i=tot;i>101+x;i--)
				a[i]=a[i-1];
			a[101+x]=++k;
		}
	cout << tot << endl;
	for (int i=1;i<=tot;i++)
		cout << a[i] << " ";
	return 0;
}
```