#include <iostream>
#include <vector>

using namespace std;

long long n, id, tot;

vector<long long> st, ed;

int main()
{
	cin >> n;
	while (n)
	{
		if (n & 1) ed.push_back(tot++), n >>= 1;
		else st.push_back(tot++), n--;
	}
	
	tot = st.size() + ed.size();
	
	cout << tot + 100 << endl;
	for (int i = 0; i < st.size(); i++) cout << tot-st[i] << " ";
	for (int i = ed.size()-1; i >= 0; i--) cout << tot-ed[i] << " ";
	for (int i = 1; i <= 100; i++) cout << i << " ";
	
	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 <iostream>
#include <vector>

using namespace std;

long long n, id, tot;

vector<long long> st, ed;

int main()
{
	cin >> n;
	while (n)
	{
		if (n & 1) ed.push_back(tot++), n >>= 1;
		else st.push_back(tot++), n--;
	}
	
	tot = st.size() + ed.size();
	
	cout << tot + 100 << endl;
	for (int i = 0; i < st.size(); i++) cout << tot-st[i] << " ";
	for (int i = ed.size()-1; i >= 0; i--) cout << tot-ed[i] << " ";
	for (int i = 1; i <= 100; i++) cout << i << " ";
	
	return 0;
}
```