#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
long long n;
int m, i, j, k, l, r, z, a[405], len;
int po[80];
int main()
{
	cin >> n;
	if (n == 1) {cout << "2\n 1 1"; return 0;}
	n ++;
	l = 200; r = 199;
	while (n != 1)
	{
		++len;
		if (n & 1) po[len] = 1;
		n >>= 1;
	}
	//cout << len << endl;
	for(i = len; i >= 1; i --)
	{
		a[++r] = ++z;
		if (po[i]) a[--l] = ++z;
	}
	cout << (r - l + 1) * 2 << endl;
	for(i = l; i <= r; i ++)
		cout << a[i] << " ";
	for(i = 1; i <= r - l + 1; i ++)
		cout << i << " ";
}
### Prompt

Construct a cpp code solution to the problem outlined:

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<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
long long n;
int m, i, j, k, l, r, z, a[405], len;
int po[80];
int main()
{
	cin >> n;
	if (n == 1) {cout << "2\n 1 1"; return 0;}
	n ++;
	l = 200; r = 199;
	while (n != 1)
	{
		++len;
		if (n & 1) po[len] = 1;
		n >>= 1;
	}
	//cout << len << endl;
	for(i = len; i >= 1; i --)
	{
		a[++r] = ++z;
		if (po[i]) a[--l] = ++z;
	}
	cout << (r - l + 1) * 2 << endl;
	for(i = l; i <= r; i ++)
		cout << a[i] << " ";
	for(i = 1; i <= r - l + 1; i ++)
		cout << i << " ";
}
```