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.io.BufferedReader
import java.io.InputStream
import java.io.InputStreamReader
import java.util.*

fun main(args: Array<String>) {
    val sc = FastScanner(System.`in`)

    val n = sc.nextInt()
    val a = Array(n) { sc.nextInt() }
    val b = Array(n) { sc.nextInt() }

    var best = -1
    for (bit in 0..(1.shl(n))) {
        val res = mutableListOf<Int>()

        var count = 0
        for (x in 0 until n) {
            if (bit.and(1.shl(x)) != 0) {
                res.add(b[x])
                count += 1
            } else res.add(a[x])
        }
        if (count % 2 != 0) continue

        val ch = check(res, bit)
        if ((best == -1 || count < best) && ch!=-1) best = ch /2
    }

    println(best)
}

fun check(cards: MutableList<Int>, bit:Int): Int {
    val flips = Array(51) { 0 }
    for (i in 0 until cards.size) {
        if(bit.and(1.shl(i)) != 0)
            flips[cards[i]] += 1
    }

    val before = Array(51) { 0 }
    for (i in 0 until cards.size) {
        before[cards[i]] += i
    }

    cards.sort()

    val after = Array(51) { 0 }
    for (i in 0 until cards.size) {
        after[cards[i]] += i
    }

    var sum = 0
    for (i in 0..50) {
        sum += Math.abs(before[i] - after[i])
        if ((before[i] + after[i] + flips[i]) % 2 !=0 ){
            return -1
        }
    }
    return sum
}


class FastScanner(s: InputStream) {
    private var st = StringTokenizer("")
    private val br = BufferedReader(InputStreamReader(s))

    fun next(): String {
        while (!st.hasMoreTokens()) st = StringTokenizer(br.readLine())

        return st.nextToken()
    }

    fun nextInt() = next().toInt()
    fun nextLong() = next().toLong()
    fun nextLine() = br.readLine()
    fun nextDouble() = next().toDouble()
    fun ready() = br.ready()
}