#pragma GCC optimize("O3")
#include<bits/stdc++.h> 
using namespace std;
using ll=long long;
using P=pair<ll,ll>;
template<class T> using V=vector<T>; 
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
const ll inf=(1e18);
const ll mod=998244353;
// const ll mod=1000000007;
const vector<int> dy={-1,0,1,0},dx={0,-1,0,1};
ll GCD(ll a,ll b) {return b ? GCD(b,a%b):a;}
ll LCM(ll c,ll d){return c/GCD(c,d)*d;}
struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(20);}} __init;
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
template<class T>void debag(const vector<T> &a){cerr<<"debag :";for(auto v:a)cerr<<v<<" ";cerr<<"\n";}
template<class T>void print(const vector<T> &a){for(auto v:a)cout<<v<<" ";cout<<"\n";}
int main(){
      int n;
      cin>>n;
      if(n%3==0){
            V<int> a(n),b(n);
            for(int i=0;i<n;i++){
                  a[i]=i*3;
                  b[i]=i*3+1;
            }
            ll rem=n;
            for(int i=n-1;i>=0;i--){
                  if(rem==0)break;
                  a[i]+=3;
                  rem-=3;
            }
            sort(all(a));sort(all(b));
            print(a);
            print(b);
      }else if(n%3==1){
            n--;
            V<int> a(n),b(n);
            for(int i=0;i<n;i++){
                  a[i]=(i+1)*3;
                  b[i]=(i+1)*3-1;
            }
            ll rem=n-3;
            for(int i=n-1;i>=0;i--){
                  if(rem==0)break;
                  b[i]+=3;
                  rem-=3;
            }
            b.emplace_back(a.back()+6);
            a.emplace_back(a.back()+3);
            sort(all(a));sort(all(b));
            print(a);
            print(b); 
      }else{
            V<int> a(n),b(n);
            for(int i=0;i<n;i++){
                  a[i]=(i+2)*3;
                  b[i]=(i+2)*3+1;
            }  
            a[n-1]=1;a[n-2]=4;
            ll rem=n-2+(b[n-1]-a[n-1])+(b[n-2]-a[n-2]);
            sort(all(a));
            for(int i=n-1;i>=0;i--){
                  if(rem==0)break;
                  while(a[i]+3<=(i+1==n?10000:a[i+1]-1)&&rem>0){
                        a[i]+=3;
                        rem-=3;
                  }
            }
            sort(all(a));sort(all(b));
            print(a);
            print(b); 
      }
}