#include <deque>
#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long LL;

deque<int> a, b;

int Solve(LL n)
{
	if (n == 1)
	{
		a.push_front(1);
		b.push_front(1);
		return 1;
	}
	if (n & 1)
	{
		int p = Solve(n >> 1) + 1;
		a.push_back(p);
		b.push_back(p);
		return p;
	}
	else
	{
		int p = Solve(n - 1) + 1;
		a.push_front(p);
		b.push_back(p);
		return p;
	}
}

int main()
{
	LL n;
	scanf("%lld", &n);
	Solve(n);
	printf("%d\n", a.size() + b.size());
	for (; !a.empty(); a.pop_front()) printf("%d ", a.front());
	for (; !b.empty(); b.pop_front()) printf("%d ", b.front());
	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 <deque>
#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long LL;

deque<int> a, b;

int Solve(LL n)
{
	if (n == 1)
	{
		a.push_front(1);
		b.push_front(1);
		return 1;
	}
	if (n & 1)
	{
		int p = Solve(n >> 1) + 1;
		a.push_back(p);
		b.push_back(p);
		return p;
	}
	else
	{
		int p = Solve(n - 1) + 1;
		a.push_front(p);
		b.push_back(p);
		return p;
	}
}

int main()
{
	LL n;
	scanf("%lld", &n);
	Solve(n);
	printf("%d\n", a.size() + b.size());
	for (; !a.empty(); a.pop_front()) printf("%d ", a.front());
	for (; !b.empty(); b.pop_front()) printf("%d ", b.front());
	return 0;
}

```