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:
# aが1〜Nの整数、bがN〜1の整数だったら、和は全て同じ。
# そこから、例えば5番目だけ大きくしたいなら、a5〜aNとb1〜b5までにtを足せば良い。a5+b5は2t増えて、他はt増える。
# 最大にしたいところにN足して、次に大きくしたいところにN-1足して、以下同様。
# この「ある区間に一定要素を足す」は……imosすればよい。最初と最後だけメモっておいて最後に累積和。

n = int(input())
indices = list(map(int, input().split()))

a_diff = [1] * n # 1〜Nの差分をここに書いちゃう
b_diff = [-1] * n

for plus, idx in enumerate(indices):
    idx -= 1
    a_diff[idx] += plus
    # a_diff[最終要素の次] -= plus は不要。

    b_diff[0] += plus
    if idx +1 < n:
        b_diff[idx+1] -= plus

a = []
now = 1
for i in range(n):
    now += a_diff[i]
    a.append(now)

b = []
now = n+1
for i in range(n):
    now += b_diff[i]
    b.append(now)

print(' '.join(map(str, a)))
print(' '.join(map(str, b)))