#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

void solve(ll n, vector<int> &a){
	if (n == 1)
		return;
	if (n & 1)
		solve(n - 1, a), a.insert(a.begin(), a.size() + 1);
	else
		solve(n >> 1, a), a.push_back(a.size() + 1);
}

int main(){
	long long n;
	scanf("%lld", &n);
	vector<int> ans;
	solve(n + 1, ans);
	int m = ans.size();
	for (int i = 1; i <= m; ++i)
		ans.push_back(i);

	printf("%d\n", ans.size());
	for (int i = 0; i < ans.size(); ++i)
		printf("%d%c", ans[i], "\n "[i + 1 < ans.size()]);
	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;

typedef long long ll;

void solve(ll n, vector<int> &a){
	if (n == 1)
		return;
	if (n & 1)
		solve(n - 1, a), a.insert(a.begin(), a.size() + 1);
	else
		solve(n >> 1, a), a.push_back(a.size() + 1);
}

int main(){
	long long n;
	scanf("%lld", &n);
	vector<int> ans;
	solve(n + 1, ans);
	int m = ans.size();
	for (int i = 1; i <= m; ++i)
		ans.push_back(i);

	printf("%d\n", ans.size());
	for (int i = 0; i < ans.size(); ++i)
		printf("%d%c", ans[i], "\n "[i + 1 < ans.size()]);
	return 0;
}

```