#include<bits/stdc++.h>
#define ll long long
using namespace std;

int n,a[205],L,R,now;
ll cnt;

inline void solve(){
	int i=40; L=101,R=L-1;
	for(;!((1ll<<i)&cnt);i--);
	for(i--;i>=0;i--){
		a[++R]=++now;
		if((1ll<<i)&cnt) a[--L]=++now;
	}
	
	n=R-L+1;
	for(int i=1;i<=n;i++) a[i]=a[i+L-1];
	for(int i=n+1;i<=n*2;i++) a[i]=i-n;
	n<<=1;
}

int main(){
	scanf("%lld",&cnt),cnt++;
	solve(),printf("%d\n",n);
	for(int i=1;i<=n;i++) printf("%d ",a[i]);
	puts("");
	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<bits/stdc++.h>
#define ll long long
using namespace std;

int n,a[205],L,R,now;
ll cnt;

inline void solve(){
	int i=40; L=101,R=L-1;
	for(;!((1ll<<i)&cnt);i--);
	for(i--;i>=0;i--){
		a[++R]=++now;
		if((1ll<<i)&cnt) a[--L]=++now;
	}
	
	n=R-L+1;
	for(int i=1;i<=n;i++) a[i]=a[i+L-1];
	for(int i=n+1;i<=n*2;i++) a[i]=i-n;
	n<<=1;
}

int main(){
	scanf("%lld",&cnt),cnt++;
	solve(),printf("%d\n",n);
	for(int i=1;i<=n;i++) printf("%d ",a[i]);
	puts("");
	return 0; 
}
```