#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

int ans[500];
bool in[500];

int main() {
  ll n=0;
  scanf("%lld",&n);
  int cnt=0;
  while (n)
  	if (n&1) {
  		cnt++;
  		in[cnt]=0;
  		n>>=1;
	  }
	else {
	    cnt++;
	    in[cnt]=1;
	    n--;
	}
  int l=251,r=250;
  for(int i=cnt;i>0;i--)
    if (in[i]) ans[--l]=cnt-i+1; else ans[++r]=cnt-i+1;
  printf("%d\n",2*cnt);
  for(int i=l;i<=r;i++) printf("%d ",ans[i]);
  for(int i=1;i<=cnt;i++) printf("%d ",i);
  printf("\n");
  return 0;
}
### Prompt

In cpp, your task is to solve 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;

int ans[500];
bool in[500];

int main() {
  ll n=0;
  scanf("%lld",&n);
  int cnt=0;
  while (n)
  	if (n&1) {
  		cnt++;
  		in[cnt]=0;
  		n>>=1;
	  }
	else {
	    cnt++;
	    in[cnt]=1;
	    n--;
	}
  int l=251,r=250;
  for(int i=cnt;i>0;i--)
    if (in[i]) ans[--l]=cnt-i+1; else ans[++r]=cnt-i+1;
  printf("%d\n",2*cnt);
  for(int i=l;i<=r;i++) printf("%d ",ans[i]);
  for(int i=1;i<=cnt;i++) printf("%d ",i);
  printf("\n");
  return 0;
}
```