#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 7;
const int MOD = 1e9 + 7;

deque<int> ans;

void build(ll a)
{
	if (a == 1)
		return;
	build(a >> 1);
	ans.push_back(ans.size());
	if (a & 1)
		ans.push_front(ans.size());
}


int main()
{
	ios::sync_with_stdio(false);

	ll n;
	cin >> n;
	build(n + 1);
	int sz = ans.size();
	for (int i=0; i<sz; i++)
		ans.push_back(i);
	cout << 2 * sz << "\n";
	for (int i=0; i<2*sz; i++)
		cout << ans[i]+1 << " ";
	cout << "\n";

	return 0;
}
### Prompt

Create a solution in cpp for 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>
using namespace std;
typedef long long ll;
const int N = 2e5 + 7;
const int MOD = 1e9 + 7;

deque<int> ans;

void build(ll a)
{
	if (a == 1)
		return;
	build(a >> 1);
	ans.push_back(ans.size());
	if (a & 1)
		ans.push_front(ans.size());
}


int main()
{
	ios::sync_with_stdio(false);

	ll n;
	cin >> n;
	build(n + 1);
	int sz = ans.size();
	for (int i=0; i<sz; i++)
		ans.push_back(i);
	cout << 2 * sz << "\n";
	for (int i=0; i<2*sz; i++)
		cout << ans[i]+1 << " ";
	cout << "\n";

	return 0;
}
```