#include  <bits/stdc++.h>

using namespace std;

long long n,op[120],tot;
deque <int> L;
int main()
{
	scanf("%lld",&n); n++; while(n>1)
	{
		op[++op[0]]=n&1; n>>=1;
	}tot=1;
//	for(int i=1;i<=op[0];i++) printf("%d ",op[i]); printf("\n");
	for(int i=op[0];i>=1;i--)
		if(op[i]) L.push_back(tot++),L.push_front(tot++);
		else L.push_back(tot++);
	printf("%lld\n",tot+tot-2);
	for(int i=1;i<tot;i++) printf("%lld ",L.front()),L.pop_front();
	for(int i=1;i<tot;i++) printf("%lld ",i);
	printf("\n");
	return 0;
} 
### Prompt

Develop a solution in CPP to the problem described below:

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;

long long n,op[120],tot;
deque <int> L;
int main()
{
	scanf("%lld",&n); n++; while(n>1)
	{
		op[++op[0]]=n&1; n>>=1;
	}tot=1;
//	for(int i=1;i<=op[0];i++) printf("%d ",op[i]); printf("\n");
	for(int i=op[0];i>=1;i--)
		if(op[i]) L.push_back(tot++),L.push_front(tot++);
		else L.push_back(tot++);
	printf("%lld\n",tot+tot-2);
	for(int i=1;i<tot;i++) printf("%lld ",L.front()),L.pop_front();
	for(int i=1;i<tot;i++) printf("%lld ",i);
	printf("\n");
	return 0;
} 
```