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: Time Limit Exceeded)

def pop_count(bits)
    bits = (bits & 0x55555555) + (bits >> 1 & 0x55555555)
    bits = (bits & 0x33333333) + (bits >> 2 & 0x33333333)
    bits = (bits & 0x0f0f0f0f) + (bits >> 4 & 0x0f0f0f0f)
    bits = (bits & 0x00ff00ff) + (bits >> 8 & 0x00ff00ff)
    return (bits & 0x0000ffff) + (bits >>16 & 0x0000ffff)
end

N = gets.to_i
A = gets.split.map(&:to_i)
B = gets.split.map(&:to_i)
INF = 10**9
dp = Array.new(1<<N){Array.new(N, INF)}
N.times do |i|
    dp[1<<i][i] = 0
end

(1<<N).times do |b|
    cnt = pop_count(b)
    N.times do |last|
        next if b[last] == 0
        v1 = ((cnt-last+1)%2 == 1 ? B[last] : A[last])
        d2 = dp[b][last] + cnt
        N.times do |nxt|
            if b[nxt] == 1
                d2 -= 1
                next
            end
            next if v1 > ((cnt-nxt)%2 == 1 ? B[nxt] : A[nxt])
            b2 = b + (1<<nxt)
            dp[b2][nxt] = d2 if dp[b2][nxt] > d2
        end
    end
end

ans = dp[(1<<N)-1].min
ans = -1 if ans == INF
puts ans