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: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <stack>
#include <tuple>
#include <cmath>
#include <iomanip>
#include <map>
#include <cstring> //memset(dp,0,sizeof(dp))
#define ll long long
#define rep(i,n) for(int i=0;i<(n);i++)
#define repp(i,n) for(int i=n-1;i>=0;i--)
#define fi first
#define se second
#define pb push_back
using namespace std;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
typedef pair<int,int> P;
typedef pair<ll,ll> Pll;
//小数点以下　cout << fixed << setprecision(10) << num << endl;
//桁数はlog10(mid)でやれ　log(mid)/log(10)はWA
//long*long<0で判定するな
//開区間と閉区間
//アルファベットが何番目かは x=s[i]-'a'
//bfsはmain()内で　rep(i,h)rep(j,w)rep(i,4)でバグ
//切り上げ：(a+(b-1))/b
//32bit以上は 1ull<<j]
// map 初期値0 for(auto itr=mp.begin();itr!=mp.end();++itr) itr->fi itr->se で全探索
// for(auto u:mp) u.fi u.se も可能
//n個のgcd,lcmは一つずつみて更新していく
const int ma=1000000000;
int main(){
    int n; cin >> n;
    int a[n],b[n];
    rep(i,n) cin >> a[i];
    rep(i,n) cin >> b[i];
    int ans=-1;
    bool all=1;
    rep(i,(1<<n)){
        vector<pair<int,pair<int,int>>> d; //fi==key se==num
        rep(j,n){
            if(i&(1<<j)){
                d.pb({a[j],{j,0}});
            }else{
                d.pb({b[j],{j,1}});
            }
        }
        sort(d.begin(),d.end());
        bool ok=1;
        rep(j,n-1){
            if(abs(j-d[j].se.fi)%2==0 && d[j].se.se==1 && d[j].fi==d[j+1].fi) swap(d[j],d[j+1]);
            else if(abs(j-d[j].se.fi)%2!=0 && d[j].se.se==0 && d[j].fi==d[j+1].fi) swap(d[j],d[j+1]);
        }
        rep(j,n){
            if(abs(j-d[j].se.fi)%2==0 && d[j].se.se==1) ok=0;
            else if(abs(j-d[j].se.fi)%2!=0 && d[j].se.se==0) ok=0;
        }
        if(ok){
            all=0;
            //cout << "ok" << endl;
            int res=0;
            rep(k,n){
                rep(j,n-2){
                    if(d[j].fi==d[j+2].fi && abs(j-d[j].se.fi)+abs(j+2-d[j+2].se.fi)>abs(j+2-d[j].se.fi)+abs(j-d[j+2].se.fi)) swap(d[j],d[j+2]);
                }
            }
            rep(j,n){
                res+=abs(j-d[j].se.fi);
            }
            res/=2;
            if(ans==-1) ans=res;
            ans=min(ans,res);
        }
    }
    if(all){
        cout << -1 << endl;
        return 0;
    }
    cout << ans << endl;
}