#include <bits/stdc++.h>
using namespace std;
#define LL long long

const int S = 205;

LL n;
int bit[60], m;
int s[S], tot;

int main() {
	cin >> n;
	for (m = -1; n > 0; n >>= 1)
		bit[++m] = n & 1;
	
	for (int i = 1; i <= m; i++) s[++tot] = i;
	for (int i = 1; i <= m; i++) s[++tot] = i;
	int Max = (s[++tot] = m+1, s[++tot] = m+1);
	
	for (int i = m-1; i >= 0; i--) if (bit[i]) {
		for (int j = ++tot; j >= i+2; j--)
			s[j] = s[j-1];
		s[i+1] = ++Max;
		s[++tot] = Max;
	}
	
	printf("%d\n", tot);
	for (int i = 1; i <= tot; i++)
		printf("%d ", s[i]);
	puts("");
	
	return 0;
}
### Prompt

Please formulate a Cpp solution to 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 LL long long

const int S = 205;

LL n;
int bit[60], m;
int s[S], tot;

int main() {
	cin >> n;
	for (m = -1; n > 0; n >>= 1)
		bit[++m] = n & 1;
	
	for (int i = 1; i <= m; i++) s[++tot] = i;
	for (int i = 1; i <= m; i++) s[++tot] = i;
	int Max = (s[++tot] = m+1, s[++tot] = m+1);
	
	for (int i = m-1; i >= 0; i--) if (bit[i]) {
		for (int j = ++tot; j >= i+2; j--)
			s[j] = s[j-1];
		s[i+1] = ++Max;
		s[++tot] = Max;
	}
	
	printf("%d\n", tot);
	for (int i = 1; i <= tot; i++)
		printf("%d ", s[i]);
	puts("");
	
	return 0;
}
```