Problem: 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.

Examples

Input

2
1 2


Output

1 4
5 4


Input

3
3 2 1


Output

1 2 3
5 3 1


Input

3
2 3 1


Output

5 10 100
100 10 1

Solution (in PYTHON3):

Solution:
# http://tutuz.hateblo.jp/entry/2019/02/25/154609

n = int(input())
p = tuple(map(int, input().split()))

a = [-1] * n
for x, ind in enumerate(p, 1):
    a[ind - 1] = x

# {a_pi+b_pi}を満たす
# {an},{bn}を満たさない

# {an}を満たす
# 累積和を取る

# 単純に累積和を取ると、{a_pi+b_pi}が崩れる
# {bn}から差し引くことで、累積和をとっても、{a_pi+b_pi}が崩れない

b = [0] * n
for i in range(1, n):
    b[i] -= a[i - 1]  # bにaの累積和をキャンセルする / これにより、すでに{bn}は単調減少
    a[i] += a[i - 1]  # 累積和

abs_min = abs(min(b)) + 1
b = list(map(lambda x: x + abs_min, b))

print(*a)
print(*b)