#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
ll c[25];int v[25];
int main(){
	ll n;
	scanf("%lld",&n);
	ll k=1;int m=1;
	while((k<<1)-1<=n){
		k<<=1;
		m++;
	}
	n-=(k-1);
//	printf("%lld\n",n);
	int l=m/2;
	c[0]=1;
	for(int i=1;i<=l;i++)c[i]=c[i-1]*(m-i+1)/i;
	int num=0;
	while(n){
		while(n>=c[l])v[l]++,n-=c[l],num++;
		l--;
	}
	printf("%d\n",m+2*num);
	int id=1;
	for(int i=0;i<m;i++){
		if(i<=m/2)for(int j=0;j<v[i];j++)printf("%d ",++id);
		printf("1 ");
	}
	while(id>1)printf("%d ",id--);
	puts("");
	return 0;
}
### Prompt

Please formulate 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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
ll c[25];int v[25];
int main(){
	ll n;
	scanf("%lld",&n);
	ll k=1;int m=1;
	while((k<<1)-1<=n){
		k<<=1;
		m++;
	}
	n-=(k-1);
//	printf("%lld\n",n);
	int l=m/2;
	c[0]=1;
	for(int i=1;i<=l;i++)c[i]=c[i-1]*(m-i+1)/i;
	int num=0;
	while(n){
		while(n>=c[l])v[l]++,n-=c[l],num++;
		l--;
	}
	printf("%d\n",m+2*num);
	int id=1;
	for(int i=0;i<m;i++){
		if(i<=m/2)for(int j=0;j<v[i];j++)printf("%d ",++id);
		printf("1 ");
	}
	while(id>1)printf("%d ",id--);
	puts("");
	return 0;
}
```