#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define _GLIBCXX_DEBUG
using namespace std;
using std::cout;
using std::cin;
using std::endl;
using ll=long long;
using ld=long double;
ll ILL=1167167167167167167;
const int INF=2100000000;
const ll mod=1e9+7;
#define rep(i,a) for (int i=0;i<a;i++)
template<class T> using _pq = priority_queue<T, vector<T>, greater<T>>;
template<class T> ll LB(vector<T> &v,T a){return lower_bound(v.begin(),v.end(),a)-v.begin();}
template<class T> ll UB(vector<T> &v,T a){return upper_bound(v.begin(),v.end(),a)-v.begin();}
template<class T> bool chmin(T &a,const T &b){if(a>b){a=b;return 1;}else return 0;}
template<class T> bool chmax(T &a,const T &b){if(a<b){a=b;return 1;}else return 0;}
template<class T> void So(vector<T> &v) {sort(v.begin(),v.end());}
template<class T> void Sore(vector<T> &v) {sort(v.begin(),v.end(),[](T x,T y){return x>y;});}
void yneos(bool a){if(a) cout<<"Yes\n"; else cout<<"No\n";}

//最大公約数
long long Gcd(long long a,long long b){
	if(b==0) return a;
	else return Gcd(b,a%b);
}

//最小公倍数
long long Lcm(long long a,long long b){
	return (a/Gcd(a,b))*b;
}
//カーマイケル数の出力
long long carmichael(long long a){
	long long ans=1,A=1;
	//2を素因数に持つときだけ場合わけ
	while(a%2==0) A*=2,a/=2;
	if(A==4) ans=2;
	else if(A>4) ans=A/4;
	A=a;
	for(long long k=3;k*k<=a;k++){
	//for(auto k:div){ //配列divをsqrt(a)以下の素数を全列挙するやつにする
		if(A%k==0){
			long long B=k-1;
			A/=k;
			while(A%k==0) A/=k,B*=k;
			ans=Lcm(ans,B);
		}
	}
	if(A!=1) ans=Lcm(ans,A-1);
	return ans;
}


void solve();
//  rainy ~ 雨に打たれて ~
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	
	int t=1;
	//cin>>t;
	rep(i,t) solve();
}

void solve(){
    int N;
    int M=1e5;
    cin>>N;
    vector<int> p(N),q(N);
    int sum=0;
    int D=((N+1)/4)*2;
    rep(i,N){
        if(i<D) p[i]=i*2+1;
        else{
            p[i]=M-(N-1-i)*2;
        }
        sum+=p[i];
        if(i==D-1){
            rep(j,N){
                q[j]=p[i]+j*2+1;
                sum-=q[j];
            }
        }
    }
    q.push_back(p[D]);
    int ind=N-1;
    while(sum){
        sum-=2;
        if(q[ind+1]==q[ind]+2) ind++;
        q[ind]+=2;
    }
    rep(i,N){
        if(i) cout<<" ";
        cout<<p[i];
    }
    cout<<"\n";
    rep(i,N){
        if(i) cout<<" ";
        cout<<q[i];
    }
    cout<<"\n";
}