#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vi = deque<int>;
using db = double;
#define pb push_back
#define pf push_front

ll n;

vi a, b;
int t;

void cal(ll n) {
	if (n == 0)
		return;
	if (n % 2 == 1) {
		cal(n / 2);
		++t;
		a.pb(t);
		b.pb(t);
	} else {
		cal(n - 1);
		++t;
		a.pf(t);
		b.pb(t);
	}
}

int main() {
	cin >> n;
	
	cal(n);
	
	vi s = a;
	for (auto i : b) s.pb(i);
	
	cout << s.size() << endl;
	for (auto i : s)
		cout << i << ' ';
	cout << endl;
	
	return 0;
}

### Prompt

In Cpp, your task is to solve 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;

using ll = long long;
using vi = deque<int>;
using db = double;
#define pb push_back
#define pf push_front

ll n;

vi a, b;
int t;

void cal(ll n) {
	if (n == 0)
		return;
	if (n % 2 == 1) {
		cal(n / 2);
		++t;
		a.pb(t);
		b.pb(t);
	} else {
		cal(n - 1);
		++t;
		a.pf(t);
		b.pb(t);
	}
}

int main() {
	cin >> n;
	
	cal(n);
	
	vi s = a;
	for (auto i : b) s.pb(i);
	
	cout << s.size() << endl;
	for (auto i : s)
		cout << i << ' ';
	cout << endl;
	
	return 0;
}

```