#include <cstdio>
#include <cstring>
#include <deque>
#define LL long long
using namespace std;

LL n;
int d[210];
deque<int> S;

int main(){
	scanf("%lld",&n);
	int m=0;
	while(n){
		if(n&1){
			n=(n-1)/2;
			d[++m]=1;
		}else{
			n--;
			d[++m]=0;
		}
	}
	for(int i=m;i>=1;i--)
		if(d[i]) S.push_back(m-i+1);
		else S.push_front(m-i+1);
	printf("%d\n",m<<1);
	for(int i=0;i<m;i++)
		printf("%d ",S[i]);
	for(int i=1;i<=m;i++)
		printf("%d ",i);
}

### Prompt

Please create a solution in cpp 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 <cstdio>
#include <cstring>
#include <deque>
#define LL long long
using namespace std;

LL n;
int d[210];
deque<int> S;

int main(){
	scanf("%lld",&n);
	int m=0;
	while(n){
		if(n&1){
			n=(n-1)/2;
			d[++m]=1;
		}else{
			n--;
			d[++m]=0;
		}
	}
	for(int i=m;i>=1;i--)
		if(d[i]) S.push_back(m-i+1);
		else S.push_front(m-i+1);
	printf("%d\n",m<<1);
	for(int i=0;i<m;i++)
		printf("%d ",S[i]);
	for(int i=1;i<=m;i++)
		printf("%d ",i);
}

```