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 <cmath>
#include <cstdlib>
#include <iostream>
#include <bitset>
#include <deque>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#include <iterator>
#include <string>
#include <chrono>
#include <random>
#include <tuple>
#include <utility>
#include <fstream>
#include <complex>

const long INF = (1l << 30);
const long LINF = (1l << 60);

long n;
long a[2][19];
long ans = LINF;

int main(){
    scanf("%ld", &n);
    for(int i = 0; i < 2; i++){
        for(int j = 0; j < n; j++){
            scanf("%ld", &a[i][j]);
        }
    }
    for(long bit = 0; bit < (1 << n); bit++){//0:front, 1:back
        std::vector<long> num(n);
        std::vector<long> ord(n);
        for(int i = 0; i < n; i++){
            if(bit & (1 << i)){
                num[i] = a[1][i];
            }else{
                num[i] = a[0][i];
            }
            ord[i] = i;
        }
        std::sort(ord.begin(), ord.end(), [&](long x, long y){
            return num[x] < num[y];
        });
        /*
        printf("%ld\n", bit);
        for(int i = 0; i < n; i++){
            printf("  %2ld", ord[i]);
        }
        printf("\n");
        for(int i = 0; i < n; i++){
            printf("  %2ld", num[ord[i]]);
        }
        printf("\n");
        //*/
        long res = 0;
        for(int i = 0; i < n; i++){
            if((std::abs(ord[i] - i) % 2 == 0 && (bit & (1 << i))) || (std::abs(ord[i] - i) % 2 == 1 && (!(bit & (1 << i))))){
                res = LINF;
            }
            for(int j = 0; j < i; j++){
                if(ord[j] > ord[i]){
                    res++;
                }
            }
        }
        ans = std::min(ans, res);
    }
    if(ans == LINF){
        printf("-1\n");
    }else{
        printf("%ld\n", ans);
    }
}