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: Accepted)

<?php
$N = intval(trim(fgets(STDIN)));
$A = array_map("intval", explode(" ", trim(fgets(STDIN))));
$B = array_map("intval", explode(" ", trim(fgets(STDIN))));
 
//座圧する・本質的ではないが、計算量がかなり際どいのでDPの長さを50から36に落とせるメリットは大きい
list($_, $fuku, $atsu) =$r= atsu(array_merge($A, $B));
foreach($A as $i => $a)$A[$i] = $atsu[$a]+1;
foreach($B as $i => $b)$B[$i] = $atsu[$b]+1;
$MAX = max($atsu)+1;
 
$dp = array_fill(0, 1<<$N, array_fill(0,51,PHP_INT_MAX));
$dp[0][0] = 0;//どの数字も選んでいない状態の最大の数字は0としてよい
for($bit = 0, $NN = 1<<$N; $bit < $NN; ++$bit){
    $consumed = countBits($bit);
    $ords = [];
    $iter = $consumed;
    for($i = 0; $i < $N; ++$i)
        if(!($bit & (1<<$i)))
            $ords[$i] = $iter++;//使われていないカードの元の位置と現在位置の組を作成
    for($s = 0; $s <= $MAX; ++$s){
        if($dp[$bit][$s] == PHP_INT_MAX)continue;
        foreach($ords as $origPosition => $currentPosition){
            $newNumber = ($origPosition - $consumed) & 1 ? $B[$origPosition] : $A[$origPosition];
            if($newNumber >= $s){//広義単調増加であればよい
                $newBits = $bit | (1<<$origPosition);
                chmin($dp[$newBits][$newNumber], $dp[$bit][$s] + abs($currentPosition - $consumed));
            }
        }
    }
}
$min = min($dp[(1<<$N)-1]);
if($min == PHP_INT_MAX){
    echo -1;
}else{
    echo $min;
}
 
/**
 * ちょっと早くBitを数える -- 18回ループするより速い。
 **/
function countBits($x){
    $con = 0;
    while ($x) {
        $x &= $x-1;
        ++$con;
    }
    return $con;
}
 
/**
 * 座圧対象の配列を渡すと以下の値を返す
 * ・圧縮された配列
 * ・復元用のMap
 * ・圧縮用のMap
 **/
function atsu($array){
    $a = array_flip($array);
    $fuku = array_flip($a);
    sort($fuku);
    $atsu = array_flip($fuku);
    foreach($array as $i => $val)$array[$i] = $atsu[$val];
    return [$array, $fuku, $atsu];
}
 
function chmax(&$a, $b){
    if($a < $b){
        $a = $b;
        return true;
    }
    return false;
}
function chmin(&$a, $b){
    if($a > $b){
        $a = $b;
        return true;
    }
    return false;
}