#include<bits/stdc++.h>
using namespace std;
deque<int> fr,se;
void solve(int num,long long v){
	if(v==0) return;
	if(v&1){
		solve(num+1,v>>1);
		fr.push_front(num),se.push_front(num);
	}else{
		solve(num+1,v-1);
		fr.push_back(num),se.push_front(num);
	}
}
long long N;
int main(){
    cin>>N;
    solve(1,N);
    printf("%d\n",fr.size()+se.size());
    while(!fr.empty()) printf("%d ",fr.front()),fr.pop_front();
    while(!se.empty()) printf("%d ",se.front()),se.pop_front();
    return 0;
}
### Prompt

Your challenge is to write 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<bits/stdc++.h>
using namespace std;
deque<int> fr,se;
void solve(int num,long long v){
	if(v==0) return;
	if(v&1){
		solve(num+1,v>>1);
		fr.push_front(num),se.push_front(num);
	}else{
		solve(num+1,v-1);
		fr.push_back(num),se.push_front(num);
	}
}
long long N;
int main(){
    cin>>N;
    solve(1,N);
    printf("%d\n",fr.size()+se.size());
    while(!fr.empty()) printf("%d ",fr.front()),fr.pop_front();
    while(!se.empty()) printf("%d ",se.front()),se.pop_front();
    return 0;
}
```