#include<iostream>

using namespace std;

long long N,M;
int i,j;
long long a[105];
long long b[105];

void solver(long long n){
	if(n==1){
		i++;
		b[i]=M--;
		return ;
	}
	if(n&1) {
		i++;
		b[i]=M--;;
		solver(n>>=1);
		
	}
	else{
		j++;
		a[j]=M--;
		solver(n-1);
	}
}

int main()
{
	long long N;
	cin>>N;
	M=100;
	solver(N);
	cout<<2*(i+j)<<endl;
	for(int m=1;m<=j;m++){
		cout<<a[m]<<" ";
	}
	for(int m=i;m>=1;m--){
		cout<<b[m]<<" ";
	}
	
//	cout<<" ! "<<i<<" ! "<<j<<"!";
	for(int m=101-i-j;m<=100;m++){
		cout<<m<<" ";
	}
	return 0;
}

### Prompt

In Cpp, your task is to solve 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>

using namespace std;

long long N,M;
int i,j;
long long a[105];
long long b[105];

void solver(long long n){
	if(n==1){
		i++;
		b[i]=M--;
		return ;
	}
	if(n&1) {
		i++;
		b[i]=M--;;
		solver(n>>=1);
		
	}
	else{
		j++;
		a[j]=M--;
		solver(n-1);
	}
}

int main()
{
	long long N;
	cin>>N;
	M=100;
	solver(N);
	cout<<2*(i+j)<<endl;
	for(int m=1;m<=j;m++){
		cout<<a[m]<<" ";
	}
	for(int m=i;m>=1;m--){
		cout<<b[m]<<" ";
	}
	
//	cout<<" ! "<<i<<" ! "<<j<<"!";
	for(int m=101-i-j;m<=100;m++){
		cout<<m<<" ";
	}
	return 0;
}

```