Problem:
Score : 400 points

Problem Statement

#nck {
width: 30px;
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}

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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int
main(int argc, char *argv[])
{
  int n, p[21000];
  scanf("%d", &n);
  int s = 30000, a[21000], b[21000];
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    a[i] = (i + 1) * s;
    b[i] = (n - i) * s;
  }
  for (int i = 0; i < n; i++) {
    b[p[i] - 1] += p[i] - 1;
  }
  for (int i = 0; i < n - 1; i++) {
    printf("%d ", a[i]);
  } printf("%d\n", a[n - 1]);
  for (int i = 0; i < n - 1; i++) {
    printf("%d ", b[i]);
  } printf("%d\n", b[n - 1]);
  
  return 0;
}