#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
typedef long long ll;
ll n;int now=0;
vector<int> ans;
void dfs(ll n){
	if(n==1) return;
	if(n&1){
		dfs(n^1);
		ans.insert(ans.begin(),++now);
		return;
	}
	dfs(n>>1);
	ans.push_back(++now);
}
int main(){
	scanf("%lld",&n);
	dfs(n+1);
	printf("%d\n",ans.size()*2);
	for(int i=0;i<ans.size();i++) printf("%d ",ans[i]);
	for(int i=1;i<=now;i++) printf((i==now)?"%d":"%d ",i);
	return 0;
}
### Prompt

Create a solution in Cpp for 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 <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
typedef long long ll;
ll n;int now=0;
vector<int> ans;
void dfs(ll n){
	if(n==1) return;
	if(n&1){
		dfs(n^1);
		ans.insert(ans.begin(),++now);
		return;
	}
	dfs(n>>1);
	ans.push_back(++now);
}
int main(){
	scanf("%lld",&n);
	dfs(n+1);
	printf("%d\n",ans.size()*2);
	for(int i=0;i<ans.size();i++) printf("%d ",ans[i]);
	for(int i=1;i<=now;i++) printf((i==now)?"%d":"%d ",i);
	return 0;
}
```