#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
int l, r, a[210], now;
ll m;
int main(){
	scanf("%lld", &m); m++; l=0; r=201; now=0;
	while(m>=2){
		if(m&1){now++; a[++l]=now; m--;}
		now++; a[--r]=now; m>>=1;
	}
	printf("%d\n", now<<1);
	for(int i=1; i<=l; i++)printf("%d ", a[i]);
	for(int i=r; i<=200; i++)printf("%d ", a[i]);
	for(int i=now; i>=2; i--)printf("%d ", i); printf("1");
	return 0;
}
### Prompt

Please create a solution in Cpp 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>
#define ll long long
using namespace std;
int l, r, a[210], now;
ll m;
int main(){
	scanf("%lld", &m); m++; l=0; r=201; now=0;
	while(m>=2){
		if(m&1){now++; a[++l]=now; m--;}
		now++; a[--r]=now; m>>=1;
	}
	printf("%d\n", now<<1);
	for(int i=1; i<=l; i++)printf("%d ", a[i]);
	for(int i=r; i<=200; i++)printf("%d ", a[i]);
	for(int i=now; i>=2; i--)printf("%d ", i); printf("1");
	return 0;
}
```