#include<algorithm>
#include<cstring>
#include<cctype>
#include<cstdio>
#define rep(i,x,y) for(int i=x; i<=y; ++i)
#define repd(i,x,y) for(int i=x; i>=y; --i)

using namespace std;
typedef long long LL;
LL n;
int ans[1000],cnt[1000],tot;

int main()
{
	scanf("%lld",&n),++n;
	while(n)
	{
		cnt[++*cnt]=n&1;
		n>>=1;
	}
	int l=100,r=99;
	ans[++r]=++tot;
	repd(i,*cnt-1,1)
	{
		if(cnt[i]) ans[--l]=++tot;
		if(i>1) ans[++r]=++tot;
	}
	printf("%d\n",tot*2);
	rep(i,l,r) printf("%d ",ans[i]);
	rep(i,1,tot) 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<algorithm>
#include<cstring>
#include<cctype>
#include<cstdio>
#define rep(i,x,y) for(int i=x; i<=y; ++i)
#define repd(i,x,y) for(int i=x; i>=y; --i)

using namespace std;
typedef long long LL;
LL n;
int ans[1000],cnt[1000],tot;

int main()
{
	scanf("%lld",&n),++n;
	while(n)
	{
		cnt[++*cnt]=n&1;
		n>>=1;
	}
	int l=100,r=99;
	ans[++r]=++tot;
	repd(i,*cnt-1,1)
	{
		if(cnt[i]) ans[--l]=++tot;
		if(i>1) ans[++r]=++tot;
	}
	printf("%d\n",tot*2);
	rep(i,l,r) printf("%d ",ans[i]);
	rep(i,1,tot) printf("%d ",i);
	puts("");
	return 0;
}
```