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 <bits/stdc++.h>
using namespace std;
using Graph = vector<vector<int> >;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define reprep(i,j,h,w) rep(i,h)rep(j,w)
#define rrep(i,m,n) for(int i = m; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define aall(x,n) (x).begin(),(x).begin()+(n)
#define VEC(type,name,n) std::vector<type> name(n);rep(i,n)std::cin >> name[i];
#define pb push_back
#define pf push_front
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define mp make_pair
#define sum accumulate
#define keta fixed<<setprecision
#define vvector(name,typ,m,n,a)vector<vector<typ> > name(m,vector<typ> (n,a))
#define vvvector(name,t,l,m,n,a) vector<vector<vector<t> > > name(l, vector<vector<t> >(m, vector<t>(n,a)));
#define vvvvector(name,t,k,l,m,n,a) vector<vector<vector<vector<t> > > > name(k,vector<vector<vector<t> > >(l, vector<vector<t> >(m, vector<t>(n,a)) ));
typedef long long ll;
const int INF = 2000000000;
const ll INF64 = 1000000000000000ll;
const ll MOD = 1000000007LL;
typedef unsigned int uint;
ll merge_cnt(vector<int> &a) {
    int n = a.size();
    if (n <= 1) { return 0; }

    ll cnt = 0;
    vector<int> b(a.begin(), a.begin()+n/2);
    vector<int> c(a.begin()+n/2, a.end());

    cnt += merge_cnt(b);
    cnt += merge_cnt(c);

    int ai = 0, bi = 0, ci = 0;
    // merge の処理
    while (ai < n) {
        if ( bi < b.size() && (ci == c.size() || b[bi] <= c[ci]) ) {
            a[ai++] = b[bi++];
        } else {
            cnt += n / 2 - bi;
            a[ai++] = c[ci++];
        }
    }
    return cnt;
}
int main(){
    int n;
    std::cin >> n;
    std::vector<int> a(n),b(n);
    rep(i,n)std::cin >> a[i];
    rep(i,n)std::cin >> b[i];
    int ans=INF;
    for(int bit=0;bit<(1<<n);bit++){
        std::vector<int> x(n),xx(n);
        std::vector<bool> xxx(n);
        int cnt=0,dum=0;

        rep(i,n){
            int dum;
            if(bit&(1<<i)){
                dum=b[i];
                cnt++;
                xxx[i]=1;
            }
            else{
                dum=a[i];
                xxx[i]=0;
            }
            x[i]=dum;
            
        }
        //std::cout << bit << std::endl;
        //rep(i,n)std::cout << x[i] << " ";
        //std::cout << std::endl;        
        if(cnt%2==1)continue;
        rep(i,n)xx[i]=x[i];
        dum=merge_cnt(x);
        bool han=1;
        
        rep(j,n){
            bool dhan=0;
            rep(i,n){
                if(!xxx[j]&&x[i]==xx[j]&&abs(j-i)%2==0)dhan=1;
                if(xxx[j]&&x[i]==xx[j]&&abs(j-i)%2==1)dhan=1;
            }
            if(!dhan)han=0;
        }
        if(han)ans=min(ans,dum);
        
    }
    if(ans==INF)std::cout << -1 << std::endl;
    else std::cout << ans << std::endl;
}