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 PHP7 (7.0.15) (judge verdict: Time Limit Exceeded)

<?php
$N = intval(trim(fgets(STDIN)));
$A = array_map("intval", explode(" ", trim(fgets(STDIN))));
$B = array_map("intval", explode(" ", trim(fgets(STDIN))));
echo solve();

function solve(){
    global $N, $A, $B;
    $dp = array_fill(0, 1<<$N, array_fill(0,51,PHP_INT_MAX));
    $dp[0][0] = 0;
    for($bit = 0; $bit < (1<<$N); ++$bit){
        $con = 0;
        for ($i = 0; $i < $N; ++$i) if ($bit & (1<<$i)) ++$con;
        $ords = [];
        $iter = $con;
        for($i = 0; $i < $N; ++$i)
            if(!($bit & (1<<$i)))
                $ords[] = [$i, $iter++];
        for($s = 0; $s<=50; $s++){
            if($dp[$bit][$s] == PHP_INT_MAX)continue;
            foreach($ords as $p){
                $ns = -1;
                if(abs($p[0]-$con) % 2 == 0){
                    $ns = $A[$p[0]];
                }else{
                    $ns = $B[$p[0]];
                }
                if($ns >= $s){
                    $dp[$bit | (1<<$p[0])][$ns] = min($dp[$bit|(1<<$p[0])][$ns], $dp[$bit][$s] + abs($p[1] - $con));
                }
            }
        }
    }
    $min = min($dp[(1<<$N)-1]);
    if($min == PHP_INT_MAX){
        return -1;
    }else{
        return $min;
    }
}