#include<bits/stdc++.h>
using namespace std;
#define MAXN	2005
#define INF	1e9
#define P	1000000007
int cnt, a[MAXN];
void work(long long n) {
	if (n == 1) {
		cnt = 0;
		return;
	}
	if (n % 2 == 0) {
		work(n / 2);
		cnt++; a[cnt] = cnt;
	} else {
		work(n - 1);
		for (int i = cnt; i >= 1; i--)
			a[i + 1] = a[i];
		a[1] = ++cnt;
	}
}
int main() {
	long long n; cin >> n; n++;
	work(n);
	cout << cnt * 2 << endl;
	for (int i = 1; i <= cnt; i++)
		cout << a[i] << ' ';
	for (int i = 1; i < cnt; i++)
		cout << i << ' ';
	cout << cnt << 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<bits/stdc++.h>
using namespace std;
#define MAXN	2005
#define INF	1e9
#define P	1000000007
int cnt, a[MAXN];
void work(long long n) {
	if (n == 1) {
		cnt = 0;
		return;
	}
	if (n % 2 == 0) {
		work(n / 2);
		cnt++; a[cnt] = cnt;
	} else {
		work(n - 1);
		for (int i = cnt; i >= 1; i--)
			a[i + 1] = a[i];
		a[1] = ++cnt;
	}
}
int main() {
	long long n; cin >> n; n++;
	work(n);
	cout << cnt * 2 << endl;
	for (int i = 1; i <= cnt; i++)
		cout << a[i] << ' ';
	for (int i = 1; i < cnt; i++)
		cout << i << ' ';
	cout << cnt << endl;
	return 0;
}
```