#include <iostream>
#include <deque>
using namespace std;
long long n;
int i;
deque<int> solve(long long n)
{
    deque<int> ret;
    if(n==1) return ret;
    if(n%2)
    {
        ret=solve(n-1);
        ret.push_front(ret.size()+1);
    }
    else
    {
        ret=solve(n/2);
        ret.push_back(ret.size()+1);
    }
    return ret;
}
int main()
{
    cin>>n;n++;
    deque<int> ans=solve(n);
    int cate=ans.size();
    for(i=1;i<=cate;i++)
        ans.push_back(i);
    cout<<ans.size()<<'\n';
    for(i=0;i<ans.size();i++)
        cout<<ans[i]<<' ';
    return 0;
}

### Prompt

Your task is to create a Cpp solution 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 <deque>
using namespace std;
long long n;
int i;
deque<int> solve(long long n)
{
    deque<int> ret;
    if(n==1) return ret;
    if(n%2)
    {
        ret=solve(n-1);
        ret.push_front(ret.size()+1);
    }
    else
    {
        ret=solve(n/2);
        ret.push_back(ret.size()+1);
    }
    return ret;
}
int main()
{
    cin>>n;n++;
    deque<int> ans=solve(n);
    int cate=ans.size();
    for(i=1;i<=cate;i++)
        ans.push_back(i);
    cout<<ans.size()<<'\n';
    for(i=0;i<ans.size();i++)
        cout<<ans[i]<<' ';
    return 0;
}

```