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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, srt, end) for (long long i = (srt); i < (long long)(end); i++)

typedef pair<ll,ll> pl;
const ll INF = 1e9;

int main(){
    ll n;
    cin >> n;
    vector<ll> a(n), b(n);
    rep(i, 0, n) cin >> a[i];
    rep(i, 0, n) cin >> b[i];
    vector<vector<ll>> dp((1<<n)+1, vector<ll>(55, INF));
    dp[0][0] = 0;
    for(int bit = 0; bit < (1<<n); bit++) {
        ll con = 0;
        for(int i = 0; i < n; i++) if(bit & (1<<i)) con++;
        vector<pl> ords;
        int iter = con;
        for(int i = 0; i < n; i++) if(!(bit & (1<<i))) ords.emplace_back(i, iter++);
        for(int s = 0; s <= 50; s++){
            if(dp[bit][s] >= INF) continue;
            for(auto p : ords) {
                ll ns = -1;
                if(abs(p.first - con) % 2 == 0) ns = a[p.first];
                else ns = b[p.first];
                if(ns >= s) dp[bit | (1<<p.first)][ns] =  min(dp[bit | (1<<p.first)][ns], dp[bit][s] + abs(p.second - con));
            }
        }
    }
    ll ans = INF;
    for(int s = 0; s <= 50; s++) ans = min(ans, dp[(1<<n)-1][s]);
    cerr << ans << endl;
    if(ans < INF) cout << ans << endl;
    else cout << -1 << endl;
    return 0;
}