#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>

typedef long long ll;
const int N=210;
std::vector<int> V;
ll n;

int main()
{
	scanf("%lld",&n),n++;
	int tot=0,flag=0;
	for(ll k=1ll<<50;k;k>>=1)
	{
		if(tot){if(flag)V.push_back(++tot);else flag=1;}
		if(n&k)V.insert(V.begin(),++tot);
	}
	printf("%d\n",tot*2);
	for(auto it:V)printf("%d ",it);
	for(int i=1;i<=tot;i++)printf("%d ",i);
	printf("\n");
	return 0;
}

### Prompt

Your challenge is to write 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<cstdio>
#include<cstring>
#include<vector>

typedef long long ll;
const int N=210;
std::vector<int> V;
ll n;

int main()
{
	scanf("%lld",&n),n++;
	int tot=0,flag=0;
	for(ll k=1ll<<50;k;k>>=1)
	{
		if(tot){if(flag)V.push_back(++tot);else flag=1;}
		if(n&k)V.insert(V.begin(),++tot);
	}
	printf("%d\n",tot*2);
	for(auto it:V)printf("%d ",it);
	for(int i=1;i<=tot;i++)printf("%d ",i);
	printf("\n");
	return 0;
}

```