#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<complex>
#define lol(i,n) for(ll i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
using namespace std;

int main(){
    ll n,k=100;cin>>n;n++;
    vector<ll> v;
    for(ll i=1;(1LL<<i)<=n;i++){
	if(n&(1LL<<(i-1))){v.push_back(k);k--;}
	v.push_back(i);
    }
    cout<<v.size()*2<<endl;
    for(auto x:v)cout<<x<<" ";
    sort(v.begin(),v.end());
    for(int i=0;i<v.size();i++){
	cout<<v[i];
	if(i==v.size()-1)cout<<endl;
	else cout<<" ";
    }
    return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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<algorithm>
#include<vector>
#include<queue>
#include<complex>
#define lol(i,n) for(ll i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
using namespace std;

int main(){
    ll n,k=100;cin>>n;n++;
    vector<ll> v;
    for(ll i=1;(1LL<<i)<=n;i++){
	if(n&(1LL<<(i-1))){v.push_back(k);k--;}
	v.push_back(i);
    }
    cout<<v.size()*2<<endl;
    for(auto x:v)cout<<x<<" ";
    sort(v.begin(),v.end());
    for(int i=0;i<v.size();i++){
	cout<<v[i];
	if(i==v.size()-1)cout<<endl;
	else cout<<" ";
    }
    return 0;
}

```