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 <cstdio>
#include <iostream>

using namespace std;

int p[20005];
int a[20005], b[20005];

int main(){
    //freopen("b3.in", "r", stdin);
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) {
        scanf("%d", &p[i]);
    }
    int lsum = 1, rsum = 1;
    for (int i = 0; i < n; ++i) {
        lsum += p[i];
        a[i] = lsum + i;
    }
    for (int i = 0; i < n; ++i) {
        rsum += p[n-1-i];
        b[n-1-i] = rsum + i;
    }
    for (int i = 0; i < n; ++i) {
        printf("%d%c", a[i], i == n-1 ? '\n' : ' ');
    }
    for (int i = 0; i < n; ++i) {
        printf("%d%c", b[i], i == n-1 ? '\n' : ' ');
    }

    return 0;
}