#include<bits/stdc++.h>
#define LL long long
using namespace std;
deque<int>q;
LL n;
int idx=0;
void dfs(LL x){
    if(!x)return;
    if(x&1){
        dfs((x-1)>>1);
        q.push_back(++idx);
    }
    else{
        dfs(x-1);
        q.push_front(++idx);
    }
}
int main()
{
    int i,j;
    cin>>n;
    dfs(n);
    deque<int>::iterator it;
    cout<<2*idx<<endl;
    for(it=q.begin();it<q.end();it++)cout<<*it<<" ";
    for(i=1;i<=idx;i++)cout<<i<<" ";
    cout<<endl;
    return 0;
}

### Prompt

In Cpp, your task is to solve 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>
#define LL long long
using namespace std;
deque<int>q;
LL n;
int idx=0;
void dfs(LL x){
    if(!x)return;
    if(x&1){
        dfs((x-1)>>1);
        q.push_back(++idx);
    }
    else{
        dfs(x-1);
        q.push_front(++idx);
    }
}
int main()
{
    int i,j;
    cin>>n;
    dfs(n);
    deque<int>::iterator it;
    cout<<2*idx<<endl;
    for(it=q.begin();it<q.end();it++)cout<<*it<<" ";
    for(i=1;i<=idx;i++)cout<<i<<" ";
    cout<<endl;
    return 0;
}

```