#include <bits/stdc++.h>
#define st first
#define nd second
#define mp make_pair
#define pb push_back
#define N 1000005
using namespace std;

typedef long long ll;

ll n, m;
deque < int > q;

void yap(ll n){
	if(n == 1)
		return;
	if(n%2){
		yap(n - 1);
		q.push_back(++m);
	}else{
		yap(n/2);
		q.push_front(++m);
	}
}

int main() {
	// freopen("in.txt", "r", stdin);
	// freopen("out.txt", "w", stdout);
	scanf("%lld",&n);
	yap(n + 1);
	printf("%lld\n", m+m);
	while(!q.empty()){
		printf("%d ", q.back());
		q.pop_back();
	}
	for(int i = 1; i <= m; i++)
		printf("%d ", i);
	return 0;
}
### Prompt

Develop a solution in CPP 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>
#define st first
#define nd second
#define mp make_pair
#define pb push_back
#define N 1000005
using namespace std;

typedef long long ll;

ll n, m;
deque < int > q;

void yap(ll n){
	if(n == 1)
		return;
	if(n%2){
		yap(n - 1);
		q.push_back(++m);
	}else{
		yap(n/2);
		q.push_front(++m);
	}
}

int main() {
	// freopen("in.txt", "r", stdin);
	// freopen("out.txt", "w", stdout);
	scanf("%lld",&n);
	yap(n + 1);
	printf("%lld\n", m+m);
	while(!q.empty()){
		printf("%d ", q.back());
		q.pop_back();
	}
	for(int i = 1; i <= m; i++)
		printf("%d ", i);
	return 0;
}
```