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

typedef long long ll;
const int maxn = 200;
int a[maxn], cnt;
int num[50], t;

inline void insert(int p, int x) {
  for (int i = cnt++; i > p; i--) a[i] = a[i - 1];
  a[p] = x;
}

int main () {
  ll n; cin >> n;
  for (int i = 0; n; i++) num[t++] = n & 1, n >>= 1;
  for (int i = 1; i < t; i++) a[cnt++] = i;
  for (int i = 1; i < t; i++) a[cnt++] = i;
  a[cnt++] = a[cnt++] = t;

  int cur = t + 1;
  for (int i = t - 2; i >= 0; i--)
    if (num[i]) insert(i, cur), a[cnt++] = cur++;

  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) printf("%d ", a[i]);
  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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn = 200;
int a[maxn], cnt;
int num[50], t;

inline void insert(int p, int x) {
  for (int i = cnt++; i > p; i--) a[i] = a[i - 1];
  a[p] = x;
}

int main () {
  ll n; cin >> n;
  for (int i = 0; n; i++) num[t++] = n & 1, n >>= 1;
  for (int i = 1; i < t; i++) a[cnt++] = i;
  for (int i = 1; i < t; i++) a[cnt++] = i;
  a[cnt++] = a[cnt++] = t;

  int cur = t + 1;
  for (int i = t - 2; i >= 0; i--)
    if (num[i]) insert(i, cur), a[cnt++] = cur++;

  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) printf("%d ", a[i]);
  return 0;
}

```