Problem:
Score : 700 points

Problem Statement

We have N cards numbered 1, 2, ..., N.
Card i (1 \leq i \leq N) has an integer A_i written in red ink on one side and an integer B_i written in blue ink on the other side.
Initially, these cards are arranged from left to right in the order from Card 1 to Card N, with the red numbers facing up.

Determine whether it is possible to have a non-decreasing sequence facing up from left to right (that is, for each i (1 \leq i \leq N - 1), the integer facing up on the (i+1)-th card from the left is not less than the integer facing up on the i-th card from the left) by repeating the operation below. If the answer is yes, find the minimum number of operations required to achieve it.

Choose an integer i (1 \leq i \leq N - 1).
Swap the i-th and (i+1)-th cards from the left, then flip these two cards.

Constraints

1 \leq N \leq 18

1 \leq A_i, B_i \leq 50 (1 \leq i \leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N
B_1 B_2 ... B_N

Output

If it is impossible to have a non-decreasing sequence, print -1.
If it is possible, print the minimum number of operations required to achieve it.

Sample Input 1

3
3 4 3
3 2 3

Sample Output 1

1

By doing the operation once with i = 1, we have a sequence [2, 3, 3] facing up, which is non-decreasing.

Sample Input 2

2
2 1
1 2

Sample Output 2

-1

After any number of operations, we have the sequence [2, 1] facing up, which is not non-decreasing.

Sample Input 3

4
1 2 3 4
5 6 7 8

Sample Output 3

0

No operation may be required.

Sample Input 4

5
28 15 22 43 31
20 22 43 33 32

Sample Output 4

-1

Sample Input 5

5
4 46 6 38 43
33 15 18 27 37

Sample Output 5

3

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

de = false
require 'set'; require 'prime'
INF=100
n = gets.chomp.to_i
aorig = gets.chomp.split.map(&:to_i)
borig = gets.chomp.split.map(&:to_i)

ans = INF
(2**(n-1)).times do |num|
  bits = num.to_s(2).split(//)
  (n -1 - bits.length).times{bits.unshift('0')}
  bc = bits.count{|bit| bit == '1'}
  next if bc >= ans

  a=aorig.dup
  b=borig.dup
  # ひっくり返し
  (n-1).times do |i|
    if bits[i] == '1'
      a[i], b[i+1] = b[i+1], a[i]
      b[i], a[i+1] = a[i+1], b[i]
    end
  end
  p [bits, bc, a] if de

  cur = 0
  flg = true
  n.times do |i|
    if cur > a[i]
      flg = false
      break
    else
      cur = a[i]
    end
  end

  if flg # OK!
    p ['ok'] if de
    ans = bc
  end
end

if ans == INF
  puts -1
else
  puts ans
end