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

long long n;
int a[1010], cnt;

int main(void) {
  scanf("%lld", &n);
  deque<int> q;
  int m = 0;
  n++;
  while((1LL << m) <= n) m++;
  m -= 2;
  while(~m) {
    q.push_back(++cnt);
    if(n >> m & 1) q.push_front(++cnt);
    m--;
  }
  int now = 0;
  while(!q.empty()) {
    a[++now] = q.front(); q.pop_front();
  }
  cout << now * 2 << endl;
  for(int i = 1; i <= now; i++)
    a[now + i] = i;
  for(int i = 1; i <= now * 2; i++)
    cout << a[i] << " ";
  cout << endl;
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;

long long n;
int a[1010], cnt;

int main(void) {
  scanf("%lld", &n);
  deque<int> q;
  int m = 0;
  n++;
  while((1LL << m) <= n) m++;
  m -= 2;
  while(~m) {
    q.push_back(++cnt);
    if(n >> m & 1) q.push_front(++cnt);
    m--;
  }
  int now = 0;
  while(!q.empty()) {
    a[++now] = q.front(); q.pop_front();
  }
  cout << now * 2 << endl;
  for(int i = 1; i <= now; i++)
    a[now + i] = i;
  for(int i = 1; i <= now * 2; i++)
    cout << a[i] << " ";
  cout << endl;
  return 0;
}

```