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

int main()
{
  long long N;
  scanf("%lld",&N);
  N++;
  int C=0;
  while((1LL<<C)<=N){
    C++;
  }
  int t=1;
  vector<int> b,f;
  for(int k=C-2;k>=0;k--){
    f.push_back(t);
    t++;
    if((N>>k)&1){
      b.push_back(t);
      t++;
    }
  }
  printf("%d\n",2*(t-1));
  for(int i=1;i<t;i++){
    printf("%d ",i);
  }
  for(int x=b.size()-1;x>=0;x--){
    printf("%d ",b[x]);
  }
  for(int x=0;x<f.size();x++){
    printf("%d ",f[x]);
  }
  putchar('\n');
  return 0;
}

### 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>
using namespace std;

int main()
{
  long long N;
  scanf("%lld",&N);
  N++;
  int C=0;
  while((1LL<<C)<=N){
    C++;
  }
  int t=1;
  vector<int> b,f;
  for(int k=C-2;k>=0;k--){
    f.push_back(t);
    t++;
    if((N>>k)&1){
      b.push_back(t);
      t++;
    }
  }
  printf("%d\n",2*(t-1));
  for(int i=1;i<t;i++){
    printf("%d ",i);
  }
  for(int x=b.size()-1;x>=0;x--){
    printf("%d ",b[x]);
  }
  for(int x=0;x<f.size();x++){
    printf("%d ",f[x]);
  }
  putchar('\n');
  return 0;
}

```