#include <bits/stdc++.h>
#define mset(a,b) memset(a,b,sizeof a)
#define mcpy(a,b) memcpy(a,b,sizeof b)
#define max(a,b) ((a)<(b)?(b):(a))
#define lb(x) ((x)&(-(x)))
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
ll n;
int cnt=0,ans[505],l,r=501;
int main(){
	cin>>n;
	while(n){
		if(n&1)ans[--r]=++cnt,n>>=1;
		else ans[++l]=++cnt,n--;
	}
	printf("%d\n",cnt<<1);
	for(int i=1;i<=l;i++)printf("%d ",cnt-ans[i]+1);
	for(int i=r;i<=500;i++)printf("%d ",cnt-ans[i]+1);
	for(int i=1;i<=cnt;i++)printf("%d ",i);
	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 <bits/stdc++.h>
#define mset(a,b) memset(a,b,sizeof a)
#define mcpy(a,b) memcpy(a,b,sizeof b)
#define max(a,b) ((a)<(b)?(b):(a))
#define lb(x) ((x)&(-(x)))
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
ll n;
int cnt=0,ans[505],l,r=501;
int main(){
	cin>>n;
	while(n){
		if(n&1)ans[--r]=++cnt,n>>=1;
		else ans[++l]=++cnt,n--;
	}
	printf("%d\n",cnt<<1);
	for(int i=1;i<=l;i++)printf("%d ",cnt-ans[i]+1);
	for(int i=r;i<=500;i++)printf("%d ",cnt-ans[i]+1);
	for(int i=1;i<=cnt;i++)printf("%d ",i);
	return 0;
}
```