#include<iostream>
#include<string>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100010;
ll n;
ll a[200], b[200];

//这题我是真的不会，直接抄了
int main()
{
	cin >> n;
	int p1 = 0, p2 = 0, m = 100;
	++n;
	while (n > 1)
	{
		if (n & 1) a[++p1] = m--, --n;
		else b[++p2] = m--, n >>= 1;
	}
	printf("%d\n", p1 + p2 << 1);
	for (int i = 1; i <= p1; ++i)
		printf("%d ", a[i]);
	for (int i = p2; i >= 1; --i)
		printf("%d ", b[i]);
	for (int i = 101 - p1 - p2; i <= 100; ++i)
		printf("%d ", i);
	puts("");
	return 0;
}
### 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<iostream>
#include<string>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100010;
ll n;
ll a[200], b[200];

//这题我是真的不会，直接抄了
int main()
{
	cin >> n;
	int p1 = 0, p2 = 0, m = 100;
	++n;
	while (n > 1)
	{
		if (n & 1) a[++p1] = m--, --n;
		else b[++p2] = m--, n >>= 1;
	}
	printf("%d\n", p1 + p2 << 1);
	for (int i = 1; i <= p1; ++i)
		printf("%d ", a[i]);
	for (int i = p2; i >= 1; --i)
		printf("%d ", b[i]);
	for (int i = 101 - p1 - p2; i <= 100; ++i)
		printf("%d ", i);
	puts("");
	return 0;
}
```