#include <bits/stdc++.h>
#define LL long long
using namespace std;
int Cnt, A[1000001];
LL N;
LL h, t;

inline void dfs (LL x)
{
	if (x == 1) {h = 1000, t = 999;return ;}
	if (x % 2) 
	{
		dfs(x - 1);
		A[--h] = ++Cnt;
	}
	else
	{
		dfs(x / 2);
		A[++t] = ++Cnt;
	}
}

int main()
{
#ifdef hk_cnyali
	freopen("C.in", "r", stdin);
	freopen("C.out", "w", stdout);
#endif
	scanf("%lld", &N);
	dfs(N + 1);
	cout<<Cnt * 2<<endl;
	for (int i = h; i <= t; ++i)
		cout<<A[i]<<" ";
	for (int i = 1; i <= Cnt; ++i)
		cout<<i<<" ";
	cout<<endl;
	return 0;
}
### Prompt

Generate 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>
#define LL long long
using namespace std;
int Cnt, A[1000001];
LL N;
LL h, t;

inline void dfs (LL x)
{
	if (x == 1) {h = 1000, t = 999;return ;}
	if (x % 2) 
	{
		dfs(x - 1);
		A[--h] = ++Cnt;
	}
	else
	{
		dfs(x / 2);
		A[++t] = ++Cnt;
	}
}

int main()
{
#ifdef hk_cnyali
	freopen("C.in", "r", stdin);
	freopen("C.out", "w", stdout);
#endif
	scanf("%lld", &N);
	dfs(N + 1);
	cout<<Cnt * 2<<endl;
	for (int i = h; i <= t; ++i)
		cout<<A[i]<<" ";
	for (int i = 1; i <= Cnt; ++i)
		cout<<i<<" ";
	cout<<endl;
	return 0;
}
```