#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#define ll long long
using namespace std;

const int N=105;

int p1,p2,m;
ll n;
int a[N],b[N];

int main(){
	int i;
	scanf("%lld",&n); n++;
	p1=p2=0; m=100;
	while (n>1){
		if (n&1ll) a[++p1]=m--,n--;
		else b[++p2]=m--,n>>=1;
	}
	printf("%d\n",(p1+p2)*2);
	for (i=1;i<=p1;i++) printf("%d ",a[i]);
	for (i=p2;i>=1;i--) printf("%d ",b[i]);
	for (i=1;i<=(p1+p2);i++) printf("%d ",m+i);
	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<iostream>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#define ll long long
using namespace std;

const int N=105;

int p1,p2,m;
ll n;
int a[N],b[N];

int main(){
	int i;
	scanf("%lld",&n); n++;
	p1=p2=0; m=100;
	while (n>1){
		if (n&1ll) a[++p1]=m--,n--;
		else b[++p2]=m--,n>>=1;
	}
	printf("%d\n",(p1+p2)*2);
	for (i=1;i<=p1;i++) printf("%d ",a[i]);
	for (i=p2;i>=1;i--) printf("%d ",b[i]);
	for (i=1;i<=(p1+p2);i++) printf("%d ",m+i);
	return 0;
}
```