#include <iostream>
#include <cstdio>
#include <deque>

using namespace std;

int cur, tot;
long long n;

deque<int> A, B;

void solve(long long n) {
	if(!n) return;
	if(n & 1) {
		solve(n >> 1);
		++cur, tot += 2;
		A.push_back(cur), B.push_back(cur);
	} else {
		solve(n - 1);
		++cur, tot += 2;
		A.push_back(cur), B.push_front(cur);
	}
	return;
}

int main() {
	scanf("%lld", &n);
	solve(n);
	printf("%d\n", tot);
	for(int i = 1; i <= tot / 2; ++i) {
		printf("%d ", A.front());
		A.pop_front();
	}
	for(int i = 1; i <= tot / 2; ++i) {
		printf("%d%c", B.front(), i == tot / 2 ? '\n' : ' ');
		B.pop_front();
	}
	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 <iostream>
#include <cstdio>
#include <deque>

using namespace std;

int cur, tot;
long long n;

deque<int> A, B;

void solve(long long n) {
	if(!n) return;
	if(n & 1) {
		solve(n >> 1);
		++cur, tot += 2;
		A.push_back(cur), B.push_back(cur);
	} else {
		solve(n - 1);
		++cur, tot += 2;
		A.push_back(cur), B.push_front(cur);
	}
	return;
}

int main() {
	scanf("%lld", &n);
	solve(n);
	printf("%d\n", tot);
	for(int i = 1; i <= tot / 2; ++i) {
		printf("%d ", A.front());
		A.pop_front();
	}
	for(int i = 1; i <= tot / 2; ++i) {
		printf("%d%c", B.front(), i == tot / 2 ? '\n' : ' ');
		B.pop_front();
	}
	return 0;
}
```