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 <vector>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <iostream>
using namespace std;

int count(int s) {
    int a = 0;
    while (s > 0) { if (s & 1) a++; s /= 2; }
    return a;
}

int check(int *c, int n, int s) {
    int d[n];
    memcpy(d, c, sizeof d);
    sort(d, d+n);
    int from[n];
    fill(from, from+n, -1);
    for (int i = 0; i < n; i++) {
        int v = c[i];
        bool ok = false;
        int p = (s >> i) & 1;
        for (int j = i + p, k = i - p; j < n || k >= 0; j += 2, k -= 2) {
            if (k >= 0 && d[k] == v && from[k] == -1) {
                from[k] = i;
                ok = true;
                // cerr << i << ' ' << j << endl;
                break;
            }
            if (j < n && d[j] == v && from[j] == -1) {
                from[j] = i;
                ok = true;
                // cerr << i << ' ' << j << endl;
                break;
            }
        }
        if (!ok) return -1;
    }
    int ans = 0;
    for (int i = 0; i < n; i++)
        for (int j = i+1; j < n; j++)
            if (from[i] > from[j]) ans++;
    return ans;
}

int main() {
    int n; cin >> n;
    int a[n], b[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) cin >> b[i];
    int ans = 1<<30;
    for (int s = 0; s < 1<<n; s++) {
        if (count(s) % 2) continue;
        int c[n];
        for (int i = 0; i < n; i++) c[i] = ((s >> i) & 1) ? b[i] : a[i];
        // cerr << "s = " << s << endl;
        // cerr << "c: "; for (int i = 0; i < n; i++) cerr << c[i] << ' '; cerr << endl;
        int r = check(c, n, s);
        if (r >= 0) {
            // cerr << "ok " << s << endl;
            ans = min(ans, r);
        }
    }
    cout << (ans == 1<<30 ? -1 : ans) << endl;
}