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)

#include <bits/stdc++.h>
#define ALL(obj) begin(obj), end(obj)
#define debug(x) cerr << #x << ": " << x << '\n'
using namespace std;
template <class T>
vector<T> make_vec(size_t a) {
    return vector<T>(a);
}
template <class T, class... Ts>
auto make_vec(size_t a, Ts... ts) {
    return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}
template <class T>
bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
    if (b < a) {
        a = b;
        return 1;
    }
    return 0;
}

using ll = long long;
using ull = unsigned long long;
const int INF = 1e9;
// const int MOD = 1e9 + 7;

int main() {
    int N;
    cin >> N;
    vector<int> p(N);
    for (int i = 0; i < N; i++) {
        cin >> p.at(i);
    }
    vector<int> a(N), b(N);
    int M = 30000;
    for (int i = 0; i < N; i++) {
        a[i] = (i * M + 1) + p[i];
    }
    for (int i = 0; i < N; i++) {
        b[i] = (N - i - 1) * M + 1;
    }
    for (int i = 0; i < N; i++) {
        cout << a[i];
        if (i != N - 1) cout << " ";
    }
    cout << endl;
    for (int i = 0; i < N; i++) {
        cout << b[i];
        if (i != N - 1) cout << " ";
    }
    cout << endl;
}