#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
#define eprintf(...) fprintf(stderr,__VA_ARGS__)

ll n;
int m,idx;
vector<int> a;

int main(){
	scanf("%lld", &n);
	ll nn = n;
	n++;
	for(int j = 40; j >= 0; j--){
		if (n >> (j + 1)) {
			++idx;
			a.insert(a.begin() + a.size() / 2, idx);
			a.push_back(idx);
		}
		if ((n >> j) != 1 && (n >> j &1)) {
			++idx;
			a.insert(a.begin(), idx);
			a.push_back(idx);
		}
	}
	printf("%d\n",a.size());
	for(int i=0; i<int(a.size()); i++) printf("%d%c", a[i], " \n"[i + 1 == a.size()]);
}
### 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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
#define eprintf(...) fprintf(stderr,__VA_ARGS__)

ll n;
int m,idx;
vector<int> a;

int main(){
	scanf("%lld", &n);
	ll nn = n;
	n++;
	for(int j = 40; j >= 0; j--){
		if (n >> (j + 1)) {
			++idx;
			a.insert(a.begin() + a.size() / 2, idx);
			a.push_back(idx);
		}
		if ((n >> j) != 1 && (n >> j &1)) {
			++idx;
			a.insert(a.begin(), idx);
			a.push_back(idx);
		}
	}
	printf("%d\n",a.size());
	for(int i=0; i<int(a.size()); i++) printf("%d%c", a[i], " \n"[i + 1 == a.size()]);
}
```