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: Accepted)

import java.lang.Integer
import java.lang.Math

fun main(args: Array<String>) {
  val INF = 1000000000
  val n = readLine()!!.toInt()
  val a = readLine()!!.split(" ").map { it.toInt() }.toIntArray()
  val b = readLine()!!.split(" ").map { it.toInt() }.toIntArray()
  val m = 1 shl n
  val dp = Array(m) { IntArray(51) { INF } }
  dp[0][0] = 0
  for (k in 0 until m) {
    for (i in 1..50) dp[k][i] = Math.min(dp[k][i],dp[k][i-1])
    val x = Integer.bitCount(k)
    for (i in 0 until n) {
      if (((k shr i) and 1) == 1) continue
      val v = if ((x-i)%2 == 0) a[i] else b[i]
      val j = k or (1 shl i)
      dp[j][v] = Math.min(dp[j][v], dp[k][v]+Integer.bitCount(k shr i))
    }
  }
  val x = dp[m-1].min()!!
  println(if (x == INF) -1 else x)
}