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 <iostream>
# include <algorithm>
# include <vector>
# include <string>
# include <set>
# include <map>
# include <cmath>
# include <iomanip>
# include <functional>
# include <utility>
# include <stack>
# include <queue>
# include <list>
# include <tuple>
# include <unordered_map>
# include <numeric>
# include <complex>
# include <bitset>
using namespace std;
using LL = long long;
using ULL = unsigned long long;
constexpr int INF = 2000000000;
constexpr int HINF = INF / 2;
constexpr double DINF = 100000000000000000.0;
constexpr long long LINF = 9223372036854775807;
constexpr long long HLINF = 4500000000000000000;
constexpr long long MMOD = 500000004;
const double PI = acos(-1);
int dx[4] = { 0,1,0,-1 }, dy[4] = { 1,0,-1,0 };
# define ALL(x)     (x).begin(),(x).end()
# define UNIQ(c)    (c).erase(unique(ALL((c))), end((c)))
# define mp         make_pair
# define eb         emplace_back
# define FOR(i,a,b) for(int i=(a);i<(b);++i)
# define REP(i,n)   FOR(i,0,n)
# define INIT       std::ios::sync_with_stdio(false);std::cin.tie(0);

int a[202020], b[202020], n,p[202020];
int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> p[i];
		p[i]--;
	}
	for (int i = 0; i < n; i++) {
		a[i] = b[n - i - 1] = 45000 * i + 1;
	}
	for (int i = 0; i < n; i++)a[p[i]] += i;
	for (int i = 0; i < n; i++) {
		cout << a[i] << (i == n - 1 ? "\n" : " ");
	}
	for (int i = 0; i < n; i++) {
		cout << b[i] << (i == n - 1 ? "\n" : " ");
	}
}