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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdint.h>
#define MAX(x,y) ((x) > (y) ? (x) : (y))
#define MIN(x,y) ((x) < (y) ? (x) : (y))
#define LL int64_t
#define REP(i,n) for (LL (i) = 0; (i) < (n); i++)
#define REPr(i,n) for (LL (i) = (n); (i) --> 0;)
#define INF 1e9
int DP[1<<18][18];
int A[18][2];

int bitcount(LL bits) {
    bits = (bits & 0x5555555555555555) + (bits >> 1 & 0x5555555555555555);
    bits = (bits & 0x3333333333333333) + (bits >> 2 & 0x3333333333333333);
    bits = (bits & 0x0f0f0f0f0f0f0f0f) + (bits >> 4 & 0x0f0f0f0f0f0f0f0f);
    bits = (bits & 0x00ff00ff00ff00ff) + (bits >> 8 & 0x00ff00ff00ff00ff);
    bits = (bits & 0x0000ffff0000ffff) + (bits >>16 & 0x0000ffff0000ffff);
    return (bits & 0x00000000ffffffff) + (bits >>32 & 0x00000000ffffffff);
}

int main() {
    int N;
    LL ans = INF;
    scanf("%d", &N);
    REP(i, N) scanf("%d", &A[i][0]);
    REP(i, N) scanf("%d", &A[i][1]);

    REP(i, 1<<N) REP(j, N) DP[i][j] = INF;
    REP(j, N) DP[1<<j][j] = 0;

    REP(i, 1<<N) {
        REP(j, N) {
            if (DP[i][j] >= INF) continue;
            int ic = bitcount(i);
            int a = A[j][(j^ic^1)&1];
            REP(k, N) {
                if (i&(1<<k) || a > A[k][(k^ic)&1]) continue;
                int t = ic - bitcount(((1<<k)-1)&i);
                DP[i|(1<<k)][k] = MIN(DP[i|(1<<k)][k], DP[i][j]+t);
            }
        }
    }

    REP(j, N) ans = MIN(ans, DP[(1<<N)-1][j]);
    if (ans >= INF) ans = -1;
    printf("%d\n", ans);
    return 0;
}