#include <cstdio>
#include <vector>

typedef long long LL;

std::vector<int> Solve(LL N) {
	if (N == 1) return std::vector<int>();
	if (N & 1) {
		auto v = Solve(N - 1);
		v.insert(v.begin(), (int)v.size() + 1);
		return v;
	}
	auto v = Solve(N / 2);
	v.push_back((int)v.size() + 1);
	return v;
}

int main() {
	LL N;
	scanf("%lld", &N);
	auto Ans = Solve(N + 1);
	printf("%d\n", 2 * (int)Ans.size());
	for (int x : Ans) printf("%d ", x);
	for (int i = 1; i <= (int)Ans.size(); ++i) printf("%d ", i);
	puts("");
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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 <cstdio>
#include <vector>

typedef long long LL;

std::vector<int> Solve(LL N) {
	if (N == 1) return std::vector<int>();
	if (N & 1) {
		auto v = Solve(N - 1);
		v.insert(v.begin(), (int)v.size() + 1);
		return v;
	}
	auto v = Solve(N / 2);
	v.push_back((int)v.size() + 1);
	return v;
}

int main() {
	LL N;
	scanf("%lld", &N);
	auto Ans = Solve(N + 1);
	printf("%d\n", 2 * (int)Ans.size());
	for (int x : Ans) printf("%d ", x);
	for (int i = 1; i <= (int)Ans.size(); ++i) printf("%d ", i);
	puts("");
	return 0;
}
```