#include<bits/stdc++.h>
using namespace std;
int p[210],q[210],cnt;
long long n;
void solve(long long n)
{
    if(n==0)
        return;
    if(~n & 1)
    {
        q[++q[0]]=cnt++;
        solve(n - 1);
    }
    else
    {
        p[++p[0]]=cnt++;
        solve(n >> 1);
    }
}
int main()
{
    cin>>n;
    solve(n);
    cout<<cnt+100<<endl;
    for(int i=1;i<=q[0];++i)
        cout<<cnt-q[i]<<" ";
    for(int i=p[0];i>0;--i)
        cout<<cnt-p[i]<<" ";
    for(int i=1;i<=100;++i)
        cout<<i<<" ";
    cout<<endl;
    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;
int p[210],q[210],cnt;
long long n;
void solve(long long n)
{
    if(n==0)
        return;
    if(~n & 1)
    {
        q[++q[0]]=cnt++;
        solve(n - 1);
    }
    else
    {
        p[++p[0]]=cnt++;
        solve(n >> 1);
    }
}
int main()
{
    cin>>n;
    solve(n);
    cout<<cnt+100<<endl;
    for(int i=1;i<=q[0];++i)
        cout<<cnt-q[i]<<" ";
    for(int i=p[0];i>0;--i)
        cout<<cnt-p[i]<<" ";
    for(int i=1;i<=100;++i)
        cout<<i<<" ";
    cout<<endl;
    return 0;
}
```