#include<bits/stdc++.h>
#include<atcoder/all>
using namespace std;
using namespace atcoder;
typedef int64_t ll;
using mint=modint998244353;

typedef long double ld;
const ll MOD=1000000007;
const ll MODA=998244353;
ll vx[4]={0,1,0,-1};
ll vy[4]={1,0,-1,0};
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
long long gcd(long long a,long long b){
    a=abs(a);
    b=abs(b);
    ll gcdmax=max(a,b);
    ll gcdmin=min(a,b);
    if(gcdmin==0)return gcdmax;
    while(true){
        if(gcdmax%gcdmin==0)break;
        else gcdmax%=gcdmin;
        swap(gcdmin,gcdmax);
    }
    return gcdmin;
}
ll pow(ll N,ll P,ll M){
    if(P==0)return 1;
    else if(P%2==0){
        ll t=pow(N,P/2,M);
        return t*t%M;
    }
    else return N*pow(N,P-1,M)%M;
}
ll pow(ll N,ll P){
    if(P==0)return 1;
    else if(P%2==0){
        ll t=pow(N,P/2);
        return t*t;
    }
    else return N*pow(N,P-1);
}

vector<ll> fac;
vector<ll> finv;
vector<ll> inv;
void COMinit(ll N,ll P){
    rep(i,N+1){
        if(i==0){
            fac.push_back(1);
            finv.push_back(1);
            inv.push_back(1);
        }
        else if(i==1){
            fac.push_back(1);
            finv.push_back(1);
            inv.push_back(1);
        }
        else{
            fac.push_back(fac.at(i-1)*i%P);
            inv.push_back(P-inv.at(P%i)*(P/i)%P);
            finv.push_back(finv.at(i-1)*inv.at(i)%P);
        }
    }
}

ll COM(ll n,ll k,ll P){
    if(n<k)return 0;
    if(n<0||k<0)return 0;
    return fac.at(n)*(finv.at(k)*finv.at(n-k)%P)%P;
}


struct UnionFind {
    vector<ll> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(ll N) : par(N) { //最初は全てが根であるとして初期化
        for(ll i = 0; i < N; i++) par[i] = i;
    }

    ll root(ll x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(ll x, ll y) { // xとyの木を併合
        ll rx = root(x); //xの根をrx
        ll ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(ll x, ll y) { // 2つのデータx, yが属する木が同じならtrueを返す
        ll rx = root(x);
        ll ry = root(y);
        return rx == ry;
    }
};

int main(){
    ll N;
    cin>>N;
    vector<ll> A(N);
    rep(i,N)cin>>A[i];
    if(N==1){
        cout<<-1<<endl;
        return 0;
    }
    if(N==2){
        if(((A[0]+1)^(A[1]+2))==((A[0]+2)^(A[1]+1))){
            cout<<1<<" "<<2<<endl;
            cout<<2<<" "<<1<<endl;
        }
        else cout<<-1<<endl;
        return 0;
    }
    vector<ll> P1(N);
    vector<ll> P2(N);
    set<ll> S;
    rep(i,N)S.insert(i+1);
    for(ll i=1;i<N;i*=2){
        vector<ll> v;
        rep(j,N){
            if((A[j]&i)==0)v.push_back(j);
        }
        if(v.size()>1){
            if(i==1){
                S.erase(2);
                S.erase(3);
                P1[v[0]]=2;
                P2[v[0]]=3;
                P1[v[1]]=3;
                P2[v[1]]=2;
            }
            else{
                S.erase(i+1);
                S.erase(1);
                P1[v[0]]=i+1;
                P2[v[0]]=1;
                P1[v[1]]=1;
                P2[v[1]]=i+1;
            }
            rep(i,N){
                if(i!=v[0]&&i!=v[1]){
                    P1[i]=(*S.begin());
                    P2[i]=P1[i];
                    S.erase(P1[i]);
                }
            }
            rep(i,N){
                cout<<P1[i]<<" ";
            }
            cout<<endl;
            rep(i,N){
                cout<<P2[i]<<" ";
            }
            cout<<endl;
            return 0;
        }
    }
    cout<<-1<<endl;
}