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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define ALL(A) (A).begin(),(A).end()
#define rep(i,n) for(int i=0;i<(n);i++)

const int inf = -1*((1<<31)+1);

using namespace std;

int main(void){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout<<fixed<<setprecision(20);
    int N,ans = inf;
    cin >> N;
    vector<int> A(N),B(N);
    rep(i,N)cin >> A[i];
    rep(i,N)cin >> B[i];
    for(int bit = 0 ;bit<(1<<N);bit++){
        //bitが1になっていたら裏を使う
        vector<int> ura(N);
        for(int i=0;i<N;i++){
            //もしも、bitのi bit目が立っていたら、そのカードは裏面を使う
            if((bit>>i)&1)ura[i] = 1;
        }
        int even = 0,odd = 0; 
        //evenは偶数番目に置いたカードの枚数
        //oddは奇数番目に置いたカードの枚数

        vector<pair<int,int>> e,o;
        //eは偶数番目に使うカードの集合
        //oは奇数番目に使うカードの集合

        for(int i=0;i<N;i++){
            //i番目のカードについて考える
            if(i%2==0){
                //偶数番目のカードについて
                if(ura[i]){
                    //もしも裏面を使う場合は
                    //奇数番目に置く必要がある
                    odd++;
                    o.push_back({B[i],i});
                    //ここのiは何番目のカードかの情報
                    //最後に条件を満たす数列を作った場合に、操作回数を求めるのに使う
                }else{
                    //偶数番目に置く必要がある
                    e.push_back({A[i],i});
                    even++;
                }
            }else{
                //奇数番目のカードについて
                if(ura[i]){
                    //もしも裏面を使う場合は偶数番目に配置する必要がある
                    e.push_back({B[i],i});
                    even++;
                }else{
                    o.push_back({A[i],i});
                    odd++;
                }
            }
        }
        if(odd!= (N/2))continue;//奇数番目に置くカードの枚数が N/2じゃなかったらその置き方はできない
        sort(ALL(e));//とりあえず条件を満たすようにソートする
        sort(ALL(o));//とりあえず条件を満たすようにソートする
        //偶数、奇数と並べていく
        int evennow = 0,oddnow = 0;
        vector<pair<int,int>> p;
        for(int i=0;i<N;i++){
            if(i%2==0){
                p.push_back(e[evennow]);
                evennow++;
            }else{
                p.push_back(o[oddnow]);
                oddnow++;
            }
        }
        bool ok = 1;
        //ここでは、pが条件を満たす数列かどうかを判別する
        for(int i=0;i<N-1;i++){
            if(p[i].first>p[i+1].first)ok = 0;
        }
        if(!ok)continue;
        //pが条件を満たす数列だった場合は、操作回数を求める
        int cnt = 0;
        for(int i=0;i<N;i++){
            for(int j=i+1;j<N;j++){
                //何か自分より右にあるindex番号が小さいやつの個数を足していくと
                //バブルソートのスワップ回数になるらしい
                //証明はまだしらない
                if(p[i].second>p[j].second)cnt++;
            }
        }
        ans = min(ans,cnt);
    }
    if(ans==inf)ans = -1;
    cout << ans << endl;
}