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)

N = gets.to_i
as = gets.split.map(&:to_i)
bs = gets.split.map(&:to_i)

abs = as.zip(bs).map.with_index do |(a, b), i|
  [i, [a, b]]
end

def dfs(cond, acc)
  if cond.size == 0
    return [acc]
  end
  i, (a, b) = cond[0]
  dfs(cond.drop(1), acc + [[i, a]]) + dfs(cond.drop(1), acc + [[i, b]])
end

conds = dfs(abs, [])
sorted = conds.map do |cond|
  cond.sort_by { |_, a| a }
end
INF = 1 << 60
ans = INF
sorted.each do |l|
  index = 0
  is_ok = l.all? do |i, x|
    ret = false
    if as[i] == bs[i]
      ret = true
    else
      if (i - index).abs.odd?
        ret = x == bs[i]
      else
        ret = x == as[i]
      end
    end
    index += 1
    ret
  end
  next unless is_ok
  sum = 0
  index = 0
  l.each do |i, x|
    sum += (index - i).abs
    index += 1
  end
  next if sum.odd?
  ans = sum / 2 if sum / 2 < ans
end
if ans == INF
  puts -1
else
  puts ans
end