#include <cstdio>
using namespace std;
 
int p[110], q[110], cnt;
long long N;
     
void solve(long long n){
    if(n<1)	return;
   	if(!(n&1)){
   		q[++q[0]]=cnt++;
   		solve(n-1);
   	}
   	else{
   		p[++p[0]]=cnt++;
   		solve(n>>1);
 	}
}
    
int main(){
  	scanf("%lld", &N);
  	solve(N);
  	printf("%d\n", cnt+100);
  	for(int i=1; i<=q[0]; i++)	printf("%d ", cnt-q[i]);
  	for(int i=p[0]; i; i--)		printf("%d ", cnt-p[i]);
   	for(int i=1; i<=100; i++)	printf("%d ", i);
 	return 0;
}
### Prompt

Your task is to create 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 <cstdio>
using namespace std;
 
int p[110], q[110], cnt;
long long N;
     
void solve(long long n){
    if(n<1)	return;
   	if(!(n&1)){
   		q[++q[0]]=cnt++;
   		solve(n-1);
   	}
   	else{
   		p[++p[0]]=cnt++;
   		solve(n>>1);
 	}
}
    
int main(){
  	scanf("%lld", &N);
  	solve(N);
  	printf("%d\n", cnt+100);
  	for(int i=1; i<=q[0]; i++)	printf("%d ", cnt-q[i]);
  	for(int i=p[0]; i; i--)		printf("%d ", cnt-p[i]);
   	for(int i=1; i<=100; i++)	printf("%d ", i);
 	return 0;
}
```