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.chomp.to_i
a = gets.chomp.split.map(&:to_i)
b = gets.chomp.split.map(&:to_i)

def swap(aa, bb, i)
  aa_1 = aa[i-1]
  aa_2 = aa[i]
  aa[i-1] = bb[i]
  aa[i] = bb[i-1]
  bb[i-1] = aa_2
  bb[i] = aa_1
  [aa, bb]
end

def check(aa)
  (N-1).times do |i|
    if aa[i+1] - aa[i] < 0
      return false
    end
  end
  true
end

if check(a)
  p 0
  exit
end

que = []
que.push([a, b, 0, 10**9])
while !que.empty?
  aa, bb, index, pre = que.shift
  if index > 35
    p -1
    exit
  end
  1.upto(N-1) do |i|
    if pre == 10**9 || (i-pre).abs == 1
      aaa, bbb = swap(aa, bb, i)
      index += 1
      if check(aaa)
        p index
        exit
      end
      que.push([aaa, bbb, index, i])
    end
  end
end
p -1