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"
#define hhh cerr<<"hhh"<<endl
#define see(x) cerr<<(#x)<<'='<<(x)<<endl
using namespace std;
typedef long long ll;
typedef pair<int,int> pr;
inline int read() {int x=0,f=1;char c=getchar();while(c!='-'&&(c<'0'||c>'9'))c=getchar();if(c=='-')f=-1,c=getchar();while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();return f*x;}

const int maxn = 4e5+7;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;

int a[18], b[18];
int p1[18], p2[18], vis[18];

int main() {
    int n=read();
    for(int i=0; i<n; ++i) a[i]=read();
    for(int i=0; i<n; ++i) b[i]=read();
    int ans=inf;
    for(int s=0; s<1<<n; ++s) {
        for(int i=0; i<n; ++i) p1[i]=p2[i]=(s>>i&1?b[i]:a[i]);
        sort(p2,p2+n);
        memset(vis,0,sizeof(vis));
        int tmp=0, f=1;
        for(int i=0; i<n; ++i) {
            int h=s>>i&1;
            int t=(s>>i&1?b[i]:a[i]);
            int pos=-1;
            for(int d=0; i-d>=0||i+d<n; ++d) {
                int j=i-d;
                if(j>=0&&!vis[j]) {
                    if(p2[j]==t&&abs(j-i)%2==h) {
                        tmp+=abs(j-i);
                        vis[j]=1;
                        pos=j;
                    }
                }
                if(pos!=-1) break;
                j=i+d;
                if(j<n&&!vis[j]) {
                    if(p2[j]==t&&abs(j-i)%2==h) {
                        tmp+=abs(j-i);
                        vis[j]=1;
                        pos=j;
                    }
                }
                if(pos!=-1) break;
            }
            if(pos==-1) { f=0; break; }
        }
        if(f) ans=min(ans,tmp);
    }
    if(ans>100000) printf("-1\n");
    else printf("%d\n", ans/2);
}