#include <bits/stdc++.h>

using namespace std;

int i;
long long n;
deque <int> ans;

int main(){
	scanf("%lld", &n);
	
	n++;
	
	long long now = 1;
	int cnt = -1;
	
	while(now <= n){
		now *= 2;
		cnt++;
	}
	
	int nxt = 0;
	
	for(i = cnt - 1; i >= 0; i--){
		nxt++;
		ans.push_back(nxt);
		if(n & (1LL << i)){
			nxt++;
			ans.push_front(nxt);
		}
	}
	
	for(i = 1; i <= nxt; i++)
	ans.push_back(i);
	
	printf("%d\n", ans.size());
	
	for(i = 0; i < ans.size(); i++){
		if(i == 0)
		printf("%d", ans[i]);
		else
		printf(" %d", ans[i]);
	}
	
	printf("\n");
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;

int i;
long long n;
deque <int> ans;

int main(){
	scanf("%lld", &n);
	
	n++;
	
	long long now = 1;
	int cnt = -1;
	
	while(now <= n){
		now *= 2;
		cnt++;
	}
	
	int nxt = 0;
	
	for(i = cnt - 1; i >= 0; i--){
		nxt++;
		ans.push_back(nxt);
		if(n & (1LL << i)){
			nxt++;
			ans.push_front(nxt);
		}
	}
	
	for(i = 1; i <= nxt; i++)
	ans.push_back(i);
	
	printf("%d\n", ans.size());
	
	for(i = 0; i < ans.size(); i++){
		if(i == 0)
		printf("%d", ans[i]);
		else
		printf(" %d", ans[i]);
	}
	
	printf("\n");
}

```