#include <iostream>
#include <vector>
#include <deque>

using namespace std;

int main()
{
    long long n, m = 1LL << 40;
    cin >> n; ++n;
    while (n < m)
    m >>= 1;
    deque<int> que;
    // que.push_back(1);
    m >>= 1;
    int i = 0;
    while (m)
    {
        que.push_back(++i);
        if (n & m)
            que.push_front(++i);
        else
        ;
            // que.push_front(++i);
        m >>= 1;
    }
    cout << 2 * i << endl;
    for (auto i : que)
    cout << i << ' ';
    for (int j = 1; j <= i; ++j)
    cout << j << ' ';
    cout << endl;
    return 0;
}
### Prompt

Please create a solution in CPP 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 <iostream>
#include <vector>
#include <deque>

using namespace std;

int main()
{
    long long n, m = 1LL << 40;
    cin >> n; ++n;
    while (n < m)
    m >>= 1;
    deque<int> que;
    // que.push_back(1);
    m >>= 1;
    int i = 0;
    while (m)
    {
        que.push_back(++i);
        if (n & m)
            que.push_front(++i);
        else
        ;
            // que.push_front(++i);
        m >>= 1;
    }
    cout << 2 * i << endl;
    for (auto i : que)
    cout << i << ' ';
    for (int j = 1; j <= i; ++j)
    cout << j << ' ';
    cout << endl;
    return 0;
}
```