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 ll = long long;
#define rep(i, n) for (int i = 0; i < (n); i++)

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 < (int)b.size() && (ci == (int)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;
    cin >> n;
    // vector<int> a(n), b(n);
    // rep(i, n) cin >> a[i];
    // rep(i, n) cin >> b[i];
    vector<pair<int, int>> ab(n), tmp(n);
    rep(i, n) {
        if (i & 1) {
            cin >> ab[i].second >> ab[i].first;
        } else {
            cin >> ab[i].first >> ab[i].second;
        }
    }
    tmp = ab;
    sort(tmp.begin(), tmp.end());
    rep(i, n - 1) {
        if (i & 1) {
            if (tmp[i].second > tmp[i + 1].first) {
                puts("-1");
                return 0;
            }

        } else {
            if (tmp[i].first > tmp[i + 1].second) {
                puts("-1");
                return 0;
            }
        }
    }

    vector<int> a(n);
    rep(i, n) {
        a[i] = ab[i].first;
    }
    cout << merge_cnt(a) << endl;
}