#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll mx,n,i,j,now,ha,len,m,b[1010],c[1010],fac[1010];
int main(){
	scanf("%lld",&n);
	fac[0]=1;
	for(i=1;i<=50;i++)fac[i]=fac[i-1]*2;
	mx=50;
	while(fac[mx]>n)mx--;
	n-=fac[mx];
	m=mx;
	//printf("%lld\n",mx);
	len=m*2;
	now=m;
	c[++ha]=now+1;
	c[++ha]=now+1;
	now++;
	len+=2;
	while(n){
		while(fac[mx]>n)mx--;
		n-=fac[mx];
		b[mx]=++now;
		c[++ha]=now;
		len+=2;
	}
	printf("%lld\n",len);
	for(i=0;i<=m;i++){
		if(i)printf("%lld ",i);
		if(b[i])printf("%lld ",b[i]);
	}
	for(i=1;i<=m;i++)printf("%lld ",i);
	for(i=1;i<=ha;i++)printf("%lld ",c[i]);
}
### 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<bits/stdc++.h>
#define ll long long
using namespace std;
ll mx,n,i,j,now,ha,len,m,b[1010],c[1010],fac[1010];
int main(){
	scanf("%lld",&n);
	fac[0]=1;
	for(i=1;i<=50;i++)fac[i]=fac[i-1]*2;
	mx=50;
	while(fac[mx]>n)mx--;
	n-=fac[mx];
	m=mx;
	//printf("%lld\n",mx);
	len=m*2;
	now=m;
	c[++ha]=now+1;
	c[++ha]=now+1;
	now++;
	len+=2;
	while(n){
		while(fac[mx]>n)mx--;
		n-=fac[mx];
		b[mx]=++now;
		c[++ha]=now;
		len+=2;
	}
	printf("%lld\n",len);
	for(i=0;i<=m;i++){
		if(i)printf("%lld ",i);
		if(b[i])printf("%lld ",b[i]);
	}
	for(i=1;i<=m;i++)printf("%lld ",i);
	for(i=1;i<=ha;i++)printf("%lld ",c[i]);
}
```