#include<iostream>
#include<deque>

using namespace std;
using ll = long long;


deque<int> ans1,ans2;

void solve(ll n,int& x){
    if(n==0)return ;
    if(n==1){
        ans1.push_front(x);ans2.push_front(x);
        ++x;
        return;
    }
    solve((n-1)/2,x);
    ans1.push_front(x);ans2.push_front(x);
    ++x;
    if(n%2==0){
        ans1.push_front(x);ans2.push_back(x);
        ++x;
    }
}
int main(){
    ll n;cin>>n;
    int x = 1;
    solve(n,x);
    cout<<ans1.size()+ans2.size()<<endl;
    for(auto e : ans1)cout<<e<<" ";
    for(auto e : ans2)cout<<e<<" ";
    return 0;
}
### Prompt

Your challenge is to write 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;
using ll = long long;


deque<int> ans1,ans2;

void solve(ll n,int& x){
    if(n==0)return ;
    if(n==1){
        ans1.push_front(x);ans2.push_front(x);
        ++x;
        return;
    }
    solve((n-1)/2,x);
    ans1.push_front(x);ans2.push_front(x);
    ++x;
    if(n%2==0){
        ans1.push_front(x);ans2.push_back(x);
        ++x;
    }
}
int main(){
    ll n;cin>>n;
    int x = 1;
    solve(n,x);
    cout<<ans1.size()+ans2.size()<<endl;
    for(auto e : ans1)cout<<e<<" ";
    for(auto e : ans2)cout<<e<<" ";
    return 0;
}
```