#include<bits/stdc++.h>
using namespace std;
#define ll long long
deque<int> dq;
 
void fun(ll x){
	if(x == 1) return;
	int now;
	if(x&1LL){
		fun(x-1);
		now = dq.size() + 1;
		dq.push_front(now);
	}
	else{
		fun(x>>1);
		now = dq.size() + 1;
		dq.push_back(now);
	}
}
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	ll i,n,k,j;
	cin >> n;
	fun(n+1);
	cout << dq.size() * 2 << endl;
	for(int x: dq) 
		cout << x << " " ;
	int cnt = 1;
	while(cnt <= dq.size()) 
		cout << cnt++ << " " ;
	cout << endl;
 	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>
using namespace std;
#define ll long long
deque<int> dq;
 
void fun(ll x){
	if(x == 1) return;
	int now;
	if(x&1LL){
		fun(x-1);
		now = dq.size() + 1;
		dq.push_front(now);
	}
	else{
		fun(x>>1);
		now = dq.size() + 1;
		dq.push_back(now);
	}
}
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	ll i,n,k,j;
	cin >> n;
	fun(n+1);
	cout << dq.size() * 2 << endl;
	for(int x: dq) 
		cout << x << " " ;
	int cnt = 1;
	while(cnt <= dq.size()) 
		cout << cnt++ << " " ;
	cout << endl;
 	return 0;
} 
```