#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std;

typedef long long ll;

int tot = 0;
int Q[1010] = {};
int L = 501, R = 500;

void work(ll x) {
	if (x == 1) {
		return;
	}
	if (x & 1) {
		work(x - 1); 
		Q[--L] = ++tot;
	}
	else {
		work(x >> 1);
		Q[++R] = ++tot;
	}
}

ll N;

int main()
{
	ios::sync_with_stdio(0);
	cin >> N;
	work(N + 1);
	cout << 2 * tot << endl;
	for (int i = L; i <= R; ++i) {
		cout << Q[i] << " ";
	}
	for (int i = 1; i <= tot; ++i) {
		cout << i << " ";
	}
	cout << endl;
	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 <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std;

typedef long long ll;

int tot = 0;
int Q[1010] = {};
int L = 501, R = 500;

void work(ll x) {
	if (x == 1) {
		return;
	}
	if (x & 1) {
		work(x - 1); 
		Q[--L] = ++tot;
	}
	else {
		work(x >> 1);
		Q[++R] = ++tot;
	}
}

ll N;

int main()
{
	ios::sync_with_stdio(0);
	cin >> N;
	work(N + 1);
	cout << 2 * tot << endl;
	for (int i = L; i <= R; ++i) {
		cout << Q[i] << " ";
	}
	for (int i = 1; i <= tot; ++i) {
		cout << i << " ";
	}
	cout << endl;
	return 0;
}
```