#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
using namespace std;

typedef long long ll;

const int MAX_N = 1e12 + 12;
ll n;
vector <int> vec, t;

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    cin >> n;
    int l = 1, r = 100;

    while (n) {
        n--;
        if (n & 1)
            vec.push_back(r--);
        else {
            n >>= 1;
            vec.push_back(l++);
        }
    }

    cout << vec.size() * 2 << "\n";
    t = vec;
    sort(t.begin(), t.end());
    for (int a: t)
        cout << a << " ";
    for (int a: vec)
        cout << a << " ";
    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 <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
using namespace std;

typedef long long ll;

const int MAX_N = 1e12 + 12;
ll n;
vector <int> vec, t;

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    cin >> n;
    int l = 1, r = 100;

    while (n) {
        n--;
        if (n & 1)
            vec.push_back(r--);
        else {
            n >>= 1;
            vec.push_back(l++);
        }
    }

    cout << vec.size() * 2 << "\n";
    t = vec;
    sort(t.begin(), t.end());
    for (int a: t)
        cout << a << " ";
    for (int a: vec)
        cout << a << " ";
    return 0;
}

```