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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>) {
    keyence2020d()
}

fun keyence2020d() {
    val n = readLine()!!.toInt()
    val aList = readLine()!!.split(' ').map { it.toInt() }.toIntArray()
    val bList = readLine()!!.split(' ').map { it.toInt() }.toIntArray()

    val inf = Int.MAX_VALUE / 10
    var answer = inf

    for (rev in 0L..3L) {
        for (mid in 0 until n) {
            for (i in 0L until (1L shl (n - 1))) {
                val bitSet = BitSet.valueOf(longArrayOf(i))
                val front = aList.copyOf()
                val back = bList.copyOf()
                val revBit = BitSet.valueOf(longArrayOf(rev))

                val range1 = if (revBit[0]) 0 until mid else mid - 1 downTo 0
                val range2 = if (revBit[1]) mid..n - 2 else n - 2 downTo mid

                (range1).forEach { if (bitSet[it]) swap(front, back, it) }
                (range2).forEach { if (bitSet[it]) swap(front, back, it) }

                if ((0 until n - 1).all { front[it] <= front[it + 1] }) {
                    answer = Math.min(answer, bitSet.cardinality())
                }
            }
        }
    }

    if (answer == inf) answer = -1

    println(answer)
}

private fun swap(arr1: IntArray, arr2: IntArray, idx: Int) {
    val tempFront = arr1[idx]
    val tempBack = arr2[idx]
    arr1[idx] = arr2[idx + 1]
    arr2[idx] = arr1[idx + 1]
    arr1[idx + 1] = tempBack
    arr2[idx + 1] = tempFront
}