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: Wrong Answer)

#define _USE_MATH_DEFINES
#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
using namespace std;

//
#define int long long
#define pb(x) push_back(x)
#define m0(x) memset((x), 0, sizeof(x))
#define mm(x) memset((x), -1, sizeof(x))

//container
#define ALL(x) (x).begin(), (x).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define EACH(i, c) for (typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)
#define EXIST(s, e) ((s).find(e) != (s).end())
#define UNIQUE(v) (v).erase(unique((v).begin(), (v).end()), (v).end());
#define PERM(c) \
  sort(ALL(c)); \
  for (bool c##p = 1; c##p; c##p = next_permutation(ALL(c)))

// debug
#define GET_VAR_NAME(variable) #variable
#define test(x) cout << GET_VAR_NAME(x) << " = " << x << endl;

// bit_macro
#define bit(n) (1LL << (n))
#define bitset(a, b) (a) |= (1 << (b))
#define bitunset(a, b) (a) |= ~(1 << (b))
#define bitcheck(a, b) ((((a) >> (b)) & 1) == 1)
#define bitcount(a) __builtin_popcountll((a))

//typedef
typedef long long lint;
typedef unsigned long long ull;
typedef complex<long double> Complex;
typedef pair<int, int> P;
typedef tuple<int, int, int> TP;
typedef vector<int> vec;
typedef vector<vec> mat;

//constant
const int INF = (int)1e18;
const int MOD = (int)1e9 + 7;
const double EPS = (double)1e-10;
const int dx[] = {-1, 0, 0, 1, 0, -1, -1, 1, 1};
const int dy[] = {0, -1, 1, 0, 0, 1, -1, 1, -1};

//
template <typename T>
void chmax(T &a, T b) { a = max(a, b); }
template <typename T>
void chmin(T &a, T b) { a = min(a, b); }
//
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}

//
struct Accelerate_Cin {
  Accelerate_Cin() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(20);
  };
};

signed main() {
  int N;
  cin >> N;
  vec p(N);
  for (int i = 0; i < N; i++) cin >> p[i];

  int n = 50000;
  vec a(202020), b(202020);
  for (int i = 0; i < N; i++) {
    a[i] = n * i + 1;
    b[i] = n * (N - i) + p[i] + 1;
  }

  for (int i = 0; i < N; i++) {
    cout << a[i] << " ";
  }
  cout << endl;
  for (int i = 0; i < N; i++) {
    cout << b[i] << " ";
  }
  cout << endl;

  return 0;
}