#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;

vector<int>V;
int cnt=0;long long T;

void ins(long long x){
	if(x==1)return;
	ins(x/2);
	V.push_back(++cnt);
	if(x&1)V.insert(V.begin(),++cnt);
}

int main(){
	scanf("%lld",&T);T++;
	ins(T);
	printf("%d\n",cnt<<1);
	for(int i=0;i<V.size();i++){
		printf("%d ",V[i]);
	}for(int i=1;i<=cnt;i++){
		printf("%d ",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<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;

vector<int>V;
int cnt=0;long long T;

void ins(long long x){
	if(x==1)return;
	ins(x/2);
	V.push_back(++cnt);
	if(x&1)V.insert(V.begin(),++cnt);
}

int main(){
	scanf("%lld",&T);T++;
	ins(T);
	printf("%d\n",cnt<<1);
	for(int i=0;i<V.size();i++){
		printf("%d ",V[i]);
	}for(int i=1;i<=cnt;i++){
		printf("%d ",i);
	}return 0;
}
```