#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 666;
long long n;
int len;
int c[maxn];
int ans[maxn];
int main()
{
	cin >> n, ++n;
	int len = 0;
	while(n) c[++len] = n & 1, n >>= 1;
	int num = 0, begin = 200, end = 199;
	for(int i = len - 1; i >= 1; --i)
	{
		ans[++end] = ++num;
		if(c[i]) ans[--begin] = ++num;
	}
	cout << ((end - begin + 1) << 1) << endl;
	for(int i = begin; i <= end; ++i) printf("%d ", ans[i]);
	for(int i = begin; i <= end; ++i) printf("%d ", i - begin + 1);
}
### 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<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 666;
long long n;
int len;
int c[maxn];
int ans[maxn];
int main()
{
	cin >> n, ++n;
	int len = 0;
	while(n) c[++len] = n & 1, n >>= 1;
	int num = 0, begin = 200, end = 199;
	for(int i = len - 1; i >= 1; --i)
	{
		ans[++end] = ++num;
		if(c[i]) ans[--begin] = ++num;
	}
	cout << ((end - begin + 1) << 1) << endl;
	for(int i = begin; i <= end; ++i) printf("%d ", ans[i]);
	for(int i = begin; i <= end; ++i) printf("%d ", i - begin + 1);
}
```