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)

#ifdef DEBUG
#define _GLIBCXX_DEBUG
#endif

#include <bits/stdc++.h>

using namespace std;

#define mp make_pair
#define pb push_back
#define sz(s) ((int) ((s).size()))
#define eprintf(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)

#ifdef _WIN32
#define LLD "%I64d"
#else
#define LLD "%lld"
#endif

typedef long long ll;
typedef long double ld;

const int inf = (int) 1e9 + 100;
const ld eps = 1e-11;
const ld pi = acos(-1.0L);

mt19937 mrand(random_device{} ());
int rnd(int x) {
  return mrand() % x;
}

void precalc() {
}

const int maxn = 3e4 + 100;
int n;
int p[maxn];

bool read() {
  if (scanf("%d", &n) < 1) {
    return false;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", p + i);
    p[i]--;
  }
  return true;
}

int a[maxn], b[maxn];

void solve() {
  for (int i = 0; i < n; i++) {
    a[i] = 3e4 * i + 10;
    b[i] = 3e4 * (n - i) + 10;
  }
  for (int i = 0; i < n; i++) {
    a[p[i]] += i;
  }
  for (int i = 0; i < n; i++) {
    printf("%d%c", a[i], " \n"[i == n - 1]);
  }
  for (int i = 0; i < n; i++) {
    printf("%d%c", b[i], " \n"[i == n - 1]);
  }
}

int main() {
  precalc();
#ifdef DEBUG
  assert(freopen("text.in", "r", stdin));
  assert(freopen("text.out", "w", stdout));
#endif

  while (true) {
    if (!read()) {
      break;
    }
    solve();
#ifdef DEBUG
    eprintf("Time: %.18f\n", (double) clock() / CLOCKS_PER_SEC);
#endif
  }
  return 0;
}