#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll n;

int main() {
	cin >> n;
	vector<int> ans;

	ll p = 1, ct = 0;
	while (2*p - 1 <= n) {
		p *= 2;
		ct++;
	}

	for (int i = 0; i < ct; ++i)
		ans.push_back(i);
	for (int i = 0; i < ct; ++i)
		ans.push_back(i);

	n -= (p-1);
	int k = 60;

	for (ll j = (1LL<<60); j >= 1; j /= 2, k--) {
		if ((n|j) == n) {
			ct++;
			ans.push_back(ct);
			ans.insert(ans.begin() + k, ct);
		}
	}

	cout << ans.size() << endl;
	for (int i : ans)
		cout << i + 1 << ' ';
	cout << endl;
}
### Prompt

Please provide a cpp coded solution 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;
typedef long long ll;

ll n;

int main() {
	cin >> n;
	vector<int> ans;

	ll p = 1, ct = 0;
	while (2*p - 1 <= n) {
		p *= 2;
		ct++;
	}

	for (int i = 0; i < ct; ++i)
		ans.push_back(i);
	for (int i = 0; i < ct; ++i)
		ans.push_back(i);

	n -= (p-1);
	int k = 60;

	for (ll j = (1LL<<60); j >= 1; j /= 2, k--) {
		if ((n|j) == n) {
			ct++;
			ans.push_back(ct);
			ans.insert(ans.begin() + k, ct);
		}
	}

	cout << ans.size() << endl;
	for (int i : ans)
		cout << i + 1 << ' ';
	cout << endl;
}
```