// In The Name Of The Queen
#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long N;
    int tm = 0, tp = 0;
    vector < int > vec;
    scanf("%lld", &N); N ++;
    while (N > 1)
    {
        if (N & 1)
            vec.push_back(-- tm);
        vec.push_back(tp ++);
        N >>= 1;
    }
    int K = (int)vec.size();
    printf("%d\n", K * 2);
    for (int i = 1; i <= K; i ++)
        printf("%d ", i);
    for (int i : vec)
        printf("%d ", (i + K) % K + 1);
    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
// In The Name Of The Queen
#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long N;
    int tm = 0, tp = 0;
    vector < int > vec;
    scanf("%lld", &N); N ++;
    while (N > 1)
    {
        if (N & 1)
            vec.push_back(-- tm);
        vec.push_back(tp ++);
        N >>= 1;
    }
    int K = (int)vec.size();
    printf("%d\n", K * 2);
    for (int i = 1; i <= K; i ++)
        printf("%d ", i);
    for (int i : vec)
        printf("%d ", (i + K) % K + 1);
    return 0;
}
```