Problem:
Score : 400 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

You are given a permutation p of the set {1, 2, ..., N}. Please construct two sequences of positive integers a_1, a_2, ..., a_N and b_1, b_2, ..., b_N satisfying the following conditions:

1 \leq a_i, b_i \leq 10^9 for all i

a_1 < a_2 < ... < a_N

b_1 > b_2 > ... > b_N

a_{p_1}+b_{p_1} < a_{p_2}+b_{p_2} < ... < a_{p_N}+b_{p_N}

Constraints

2 \leq N \leq 20,000

p is a permutation of the set {1, 2, ..., N}

Input

The input is given from Standard Input in the following format:

N
p_1 p_2 ... p_N

Output

The output consists of two lines. The first line contains a_1, a_2, ..., a_N seperated by a space. The second line contains b_1, b_2, ..., b_N seperated by a space.

It can be shown that there always exists a solution for any input satisfying the constraints.

Sample Input 1

2
1 2

Sample Output 1

1 4
5 4

a_1 + b_1 = 6 and a_2 + b_2 = 8. So this output satisfies all conditions.

Sample Input 2

3
3 2 1

Sample Output 2

1 2 3
5 3 1

Sample Input 3

3
2 3 1

Sample Output 3

5 10 100
100 10 1

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
using datas=pair<ll,ll>;
using ddatas=pair<double,double>;
using tdata=pair<ll,datas>;
using vec=vector<ll>;
using mat=vector<vec>;
using pvec=vector<datas>;
using pmat=vector<pvec>;
#define For(i,a,b) for(i=a;i<b;i++)
#define bFor(i,a,b) for(i=a;i>=b;i--)
#define rep(i,N) For(i,0,N)
#define rep1(i,N) For(i,1,N)
#define brep(i,N) bFor(i,N-1,0)
#define all(v) (v).begin(),(v).end()
#define allr(v) (v).rbegin(),(v).rend()
#define vsort(v) sort(all(v))
#define vrsort(v) sort(allr(v))
#define endl "\n"
#define pb push_back
#define output(v) do{bool f=0;for(auto outi:v){cout<<(f?" ":"")<<outi;f=1;}cout<<endl;}while(0)
const ll mod=1000000007;
const ll inf=1LL<<60;
template<class T> inline bool chmax(T& a,T b){bool x=a<b;if(x)a=b;return x;} 
template<class T> inline bool chmin(T& a,T b){bool x=a>b;if(x)a=b;return x;} 

double distance(ddatas& x,ddatas& y){
  double a=x.first-y.first,b=x.second-y.second;
  return sqrt(a*a+b*b);
}

ll modinv(ll a) {
  ll b=mod,u=1,v=0,t;
  while(b){
    t=a/b;
    a-=t*b; swap(a,b);
    u-=t*v; swap(u,v);
  }
  return (u+mod)%mod;
}

ll moddevide(ll a,ll b){return (a*modinv(b))%mod;}

vec modncrlistp,modncrlistm;

ll modncr(ll n,ll r){
  ll i,size=modncrlistp.size();
  if(size<=n){
    modncrlistp.resize(n+1);
    modncrlistm.resize(n+1);
    if(!size){
      modncrlistp[0]=modncrlistm[0]=1;
      size++;
    }
    For(i,size,n+1){
      modncrlistp[i]=modncrlistp[i-1]*i%mod;
      modncrlistm[i]=modinv(modncrlistp[i]);
    }
  }
  return modncrlistp[n]*modncrlistm[r]%mod*modncrlistm[n-r]%mod;
}

ll modpow(ll a,ll n){
  ll res=1;
  while(n){
    if(n&1)res=res*a%mod;
    a=a*a%mod;
    n>>=1;
  }
  return res;
}

ll gcd(ll a,ll b){if(!b)return a;return (a%b==0)?b:gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

ll countdigits(ll n){
  ll ans=0;
  while(n){n/=10;ans++;}
  return ans;
}

ll sumdigits(ll n){
  ll ans=0;
  while(n){ans+=n%10;n/=10;}
  return ans;
}

int main(){
  ll i,N,x;
  cin>>N;
  vec a(N,1),b(N,1);
  rep(i,N){
    cin>>x;
    a[x-1]+=i;
    a[i]+=N*i;
    b[i]+=N*(N-1-i);
  }
  output(a);
  output(b);
  return 0;
}