#include <cstdio>
#include <vector>

using namespace std;

#define REP(i,n)   for(int i=0; i<(int)(n); i++)
#define FOR(i,b,e) for(int i=(b); i<=(int)(e); i++)

typedef long long ll;

ll N;

void solve() {
  N++;
  vector<int> vs;
  int k = 0;
  ll n = N;
  while (n > 1) {
    vs.push_back(++k);
    n >>= 1;
  }
  for (int i = k - 1; i >= 0; i--) {
    if (N >> i & 1) {
      vs.insert(vs.begin() + i, ++k);
    }
  }
  printf("%d\n", k * 2);
  REP(i, k)  printf("%d ", vs[i]);
  FOR(i, 1, k)  printf("%d%c", i, i == k ? '\n' : ' ');
}

void input() {
  scanf("%lld", &N);
}

int main() {
  input();
  solve();
  return 0;
}

### Prompt

Create a solution in CPP for 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 <vector>

using namespace std;

#define REP(i,n)   for(int i=0; i<(int)(n); i++)
#define FOR(i,b,e) for(int i=(b); i<=(int)(e); i++)

typedef long long ll;

ll N;

void solve() {
  N++;
  vector<int> vs;
  int k = 0;
  ll n = N;
  while (n > 1) {
    vs.push_back(++k);
    n >>= 1;
  }
  for (int i = k - 1; i >= 0; i--) {
    if (N >> i & 1) {
      vs.insert(vs.begin() + i, ++k);
    }
  }
  printf("%d\n", k * 2);
  REP(i, k)  printf("%d ", vs[i]);
  FOR(i, 1, k)  printf("%d%c", i, i == k ? '\n' : ' ');
}

void input() {
  scanf("%lld", &N);
}

int main() {
  input();
  solve();
  return 0;
}

```