#include <cstdio>
#include <vector>
#define ll long long
#define rep(i,j,k) for (i=j;i<=k;i++)
using namespace std;
ll n;
int i,k;
vector <int> a;
void solve(ll n,int &k)
{
	if (n==1) {a.push_back(1); k=1; return ;}
	if (n&1) {
		solve((n-1)/2,k);
		k++; a.push_back(k);
	}
	else {	
		solve(n-1,k);
		k++; a.insert(a.begin(),k);
	}
}
int main()
{
//	freopen("repeat.in","r",stdin);
//	freopen("repeat.out","w",stdout);
	scanf("%lld",&n);
	solve(n,k);
	rep(i,1,k) a.push_back(i);
	printf("%d\n",a.size());
	rep(i,0,a.size()-1) printf("%d ",a[i]);
	return 0;
}
### Prompt

Please provide a Cpp coded solution 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 <cstdio>
#include <vector>
#define ll long long
#define rep(i,j,k) for (i=j;i<=k;i++)
using namespace std;
ll n;
int i,k;
vector <int> a;
void solve(ll n,int &k)
{
	if (n==1) {a.push_back(1); k=1; return ;}
	if (n&1) {
		solve((n-1)/2,k);
		k++; a.push_back(k);
	}
	else {	
		solve(n-1,k);
		k++; a.insert(a.begin(),k);
	}
}
int main()
{
//	freopen("repeat.in","r",stdin);
//	freopen("repeat.out","w",stdout);
	scanf("%lld",&n);
	solve(n,k);
	rep(i,1,k) a.push_back(i);
	printf("%d\n",a.size());
	rep(i,0,a.size()-1) printf("%d ",a[i]);
	return 0;
}
```