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)

fun main(args: Array<String>) {
    val n = readLine()!!.toInt()
    val aList = readLine()!!.split(' ').map(String::toInt).toIntArray()
    val bList = readLine()!!.split(' ').map(String::toInt).toIntArray()

    var ans = Int.MAX_VALUE
    fun execute(start: Int, omote: Boolean, kaisu: Int, aListTemp: IntArray, bListTemp: IntArray) {
        val aListCopy = aListTemp.clone()
        val bListCopy = bListTemp.clone()
        if (start == n - 1) {
            if ((0..(n - 2)).all { aListCopy[it] <= aListCopy[it + 1] }) {
                ans = Math.min(ans, kaisu)
            }
            return
        }
        var min = if (start % 2 == if (omote) 0 else 1) {
            aListCopy[start]
        } else {
            Int.MAX_VALUE
        }
        var mini = -1
        if (omote) {
            for (i in start until n) {
                if (i % 2 != start % 2) continue
                if (min > aListCopy[i]) {
                    min = aListCopy[i]
                    mini = i
                }
            }
        } else {
            for (i in start until n) {
                if (i % 2 == start % 2) continue
                if (min > bListCopy[i]) {
                    min = bListCopy[i]
                    mini = i
                }
            }
        }
        if (mini == -1) {
            mini = start
        } else if (mini == start) {
            //do nothing
        } else {
            for (i in mini downTo start + 1) {
                val tampa = aListCopy[i - 1]
                val tampb = bListCopy[i - 1]
                aListCopy[i - 1] = bListCopy[i]
                bListCopy[i - 1] = aListCopy[i]
                aListCopy[i] = tampb
                bListCopy[i] = tampa
            }
        }
        execute(start + 1, true, kaisu + (mini - start), aListCopy, bListCopy)
        execute(start + 1, false, kaisu + (mini - start), aListCopy, bListCopy)
    }
    execute(0, true, 0, aList, bList)
    execute(0, false, 0, aList, bList)
    println(if (ans == Int.MAX_VALUE) -1 else ans)
}