#include <bits/stdc++.h>
using namespace std;

deque <int> De;
int cnt, cnt1, cnt2, front[105], back[105];
long long n;
void work(long long x)
{
	if (!x) return ;
	if (!(x & 1)) ++cnt, front[++cnt1] = cnt, work(x - 1);
	else ++cnt, back[++cnt2] = cnt, work(x >> 1);
}
int main()
{
	scanf("%lld", &n);
	work(n);
	printf("%d\n", cnt + 100);
	for (int i = 1; i <= cnt1; i++)
		printf("%d ", cnt - front[i] + 1);
	for (int i = cnt2; i >= 1; i--)
		printf("%d ", cnt - back[i] + 1);
	for (int i = 1; i <= 100; i++)
		printf("%d ", i);
	return 0;
}
### Prompt

Your task is to create 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 <bits/stdc++.h>
using namespace std;

deque <int> De;
int cnt, cnt1, cnt2, front[105], back[105];
long long n;
void work(long long x)
{
	if (!x) return ;
	if (!(x & 1)) ++cnt, front[++cnt1] = cnt, work(x - 1);
	else ++cnt, back[++cnt2] = cnt, work(x >> 1);
}
int main()
{
	scanf("%lld", &n);
	work(n);
	printf("%d\n", cnt + 100);
	for (int i = 1; i <= cnt1; i++)
		printf("%d ", cnt - front[i] + 1);
	for (int i = cnt2; i >= 1; i--)
		printf("%d ", cnt - back[i] + 1);
	for (int i = 1; i <= 100; i++)
		printf("%d ", i);
	return 0;
}
```