#include<cstdio>
#include<vector>
using namespace std;
vector<int> ans; 
int main()
{
	long long n;int i,j;
	scanf("%lld",&n);++n;
	for(i=50;i--;)if(n&(1LL<<i))break;
	for(j=1;j<=i;++j)ans.push_back(j);
	for(j=1;j<=i;++j)ans.push_back(j);
	for(j=i;j--;)if(n&(1LL<<j))ans.insert(ans.begin()+i,50+j);
	for(j=0;j<i;++j)if(n&(1LL<<j))ans.insert(ans.begin()+i-j,50+j);
	printf("%d\n",ans.size());
	for(i=0;i<ans.size();++i)printf("%d ",ans[i]);
}
### 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<cstdio>
#include<vector>
using namespace std;
vector<int> ans; 
int main()
{
	long long n;int i,j;
	scanf("%lld",&n);++n;
	for(i=50;i--;)if(n&(1LL<<i))break;
	for(j=1;j<=i;++j)ans.push_back(j);
	for(j=1;j<=i;++j)ans.push_back(j);
	for(j=i;j--;)if(n&(1LL<<j))ans.insert(ans.begin()+i,50+j);
	for(j=0;j<i;++j)if(n&(1LL<<j))ans.insert(ans.begin()+i-j,50+j);
	printf("%d\n",ans.size());
	for(i=0;i<ans.size();++i)printf("%d ",ans[i]);
}
```