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++ (GCC 9.2.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
using namespace std;
using     ll=long long;
using    vll=vector<   ll>;
using   vvll=vector<  vll>;
using  vvvll=vector< vvll>;
using vvvvll=vector<vvvll>;
constexpr ll INF = 1LL << 60;
struct Fast{ //cin,cout高速化のおまじない＋桁数指定
	Fast(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(std::numeric_limits<double>::max_digits10); }
} fast;
#define REPS(i, S, E) for (ll i = (S); i <= (E); i++)
#define REP(i, N) REPS(i, 0, (N)-1)
#define DEPS(i, S, E) for (ll i = (E); i >= (S); i--)
#define DEP(i, N) DEPS(i, 0, (N)-1)
#define EACH(e, v) for (auto&& e : v)
#define rep(i, S, E)  for (ll i = (S); i <= (E); i++)
#define dep(i, E, S)  for (ll i = (E); i >= (S); i--)
#define each(e, v) for (auto&& e : v)
#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()
template<class T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true; }return false; }
template<class T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true; }return false; }
template<class T> inline T MaxE(vector<T>&v,ll S,ll E){ T m=v[S]; rep(i,S,E)chmax(m,v[i]); return m; }
template<class T> inline T MinE(vector<T>&v,ll S,ll E){ T m=v[S]; rep(i,S,E)chmin(m,v[i]); return m; }
template<class T> inline T MaxE(vector<T> &v) { return MaxE(v,0,(ll)v.size()-1); }
template<class T> inline T MinE(vector<T> &v) { return MinE(v,0,(ll)v.size()-1); }
template<class T> inline T Sum(vector<T> &v,ll S,ll E){ T s=T(); rep(i,S,E)s+=v[i]; return s; }
template<class T> inline T Sum(vector<T> &v) { return Sum(v,0,v.size()-1); }
template<class T> inline ll sz(T &v){ return (ll)v.size(); }
template<class T> inline T POW(T a, ll n){ T r=1; for (; n>0; n>>=1, a*=a){ if (n&1)r*=a; } return r; }
inline ll POW(int a, ll n){ return POW((ll)a, n); }
inline ll CEIL(ll a, ll b){ return (a+b-1)/b; }


//vector用演算子テンプレート
template<class T> inline vector<T>& operator+=(vector<T> &a,const vector<T> &b){ for (ll i=0; i<(ll)a.size(); i++) a[i]+=b[i];  return a; }
template<class T> inline vector<T>& operator-=(vector<T> &a,const vector<T> &b){ for (ll i=0; i<(ll)a.size(); i++) a[i]-=b[i];  return a; }
template<class T> inline vector<T>& operator*=(vector<T> &a,const vector<T> &b){ for (ll i=0; i<(ll)a.size(); i++) a[i]*=b[i];  return a; }
template<class T> inline vector<T>& operator/=(vector<T> &a,const vector<T> &b){ for (ll i=0; i<(ll)a.size(); i++) a[i]/=b[i];  return a; }
template<class T> inline vector<T>& operator%=(vector<T> &a,const vector<T> &b){ for (ll i=0; i<(ll)a.size(); i++) a[i]%=b[i];  return a; }
template<class T,class S> inline vector<T>& operator+=(vector<T> &a,S b){ for (T &e: a) e+=b;  return a; }
template<class T,class S> inline vector<T>& operator-=(vector<T> &a,S b){ for (T &e: a) e-=b;  return a; }
template<class T,class S> inline vector<T>& operator*=(vector<T> &a,S b){ for (T &e: a) e*=b;  return a; }
template<class T,class S> inline vector<T>& operator/=(vector<T> &a,S b){ for (T &e: a) e/=b;  return a; }
template<class T,class S> inline vector<T>& operator%=(vector<T> &a,S b){ for (T &e: a) e%=b;  return a; }
template<class T,class S> inline vector<T> operator+(const vector<T> &a,S b){ vector<T> c=a; return c+=b; }
template<class T,class S> inline vector<T> operator-(const vector<T> &a,S b){ vector<T> c=a; return c-=b; }
template<class T,class S> inline vector<T> operator*(const vector<T> &a,S b){ vector<T> c=a; return c*=b; }
template<class T,class S> inline vector<T> operator/(const vector<T> &a,S b){ vector<T> c=a; return c/=b; }
template<class T,class S> inline vector<T> operator%(const vector<T> &a,S b){ vector<T> c=a; return c%=b; }
template<class T> inline vector<T> operator-(const vector<T> &a){ vector<T> c=a; return c*=(-1); }
template<class T> inline ostream &operator<<(ostream &os,const vector<T> &a){ for (ll i=0; i<(ll)a.size(); i++) os<<(i>0?" ":"")<<a[i];  return os; }


void solve()
{
	ll n;  cin >> n;
	vector<ll> p(n);  rep(i,0,n-1){ ll p_; cin>>p_; p[i]=p_; }
	p-=1;

	vll bs(n);
	rep(i,0,n-1)bs[p[i]]=i;

	vll a,b;
	a.push_back(bs[0]+1);
	b.push_back(-1);
	rep(i,1,n-1){
		ll d=bs[i]-bs[i-1];
		if (d>0){
			a.push_back(a.back()+d+1);
			b.push_back(b.back()-1);
		}
		else{
			a.push_back(a.back()+1);
			b.push_back(b.back()+d-1);
		}
	}
	ll mn=MinE(b);
	b+=(-mn+1);
	cout << a << '\n';
	cout << b << '\n';

}


int main(){
#if 1
	solve();
#else
	ll t;  cin >> t;
	rep(i, 0, t-1){
		solve();
	}
#endif
	return 0;
}