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)

#include<stdio.h>
#include<iostream>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<utility>
#include<string>
#include<cstring>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<iterator>
//#include<math.h>

using namespace std;
#define int long long
/*                      */    //#undef int
#define INF 1000000009
#define INFL 1000000000000000018ll
#define mod 1000000007ll
#define pb push_back
#define fi first
#define se second
#define mk make_pair
typedef pair<int, int> PA;
typedef priority_queue<int> PQ;
typedef priority_queue<PA, vector<PA>, greater<PA> >PQG;

int N, a[20020], b[20020], p;

signed main()
{
    cin >> N;
    for(int i = 0; i < N; i++){
        cin >> p;
        a[i] += 30000ll * (i + 1);
        b[i] += 30000ll * (N - i);
        a[p - 1] += i;
    }
    for(int i = 0; i < N; i++){
        printf("%lld", a[i]);
        if(N - 1 > i)printf(" ");
    }
    printf("\n");
    for(int i = 0; i < N; i++){
        printf("%lld", b[i]);
        if(N - 1 > i)printf(" ");
    }
    printf("\n");


    return 0;
}
/*


*/