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: Accepted)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define debug(x) cerr << #x << " : " << x << '\n'

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef long double ld;
typedef string str;
typedef pair<ll, ll> pll;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

const ll Mod = 1000000007LL;
const int Maxn = 2e5 + 10;
const ll Inf = 2242545357980376863LL;
const ll Log = 30;

ll a[Maxn], b[Maxn];
ll dp[(1<<18)+10][52];

int main(){
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	ll n, k, s;
	cin >> n;
	for(int i = 0; i < n; i++) cin >> a[i];
	for(int i = 0; i < n; i++) cin >> b[i];
	memset(dp, 31, sizeof dp);
	dp[0][0] = 0;
	
	int co, tt;
	for(int mk = 0; mk < (1 << n); mk++){
		for(int la = 0; la <= 50; la ++){
			int iv = 0;
			for(int ad = n - 1; ad >= 0; ad --){
				if(mk >> ad & 1){
					iv ++;
					continue;
				}
				co = ( (__builtin_popcount(mk) % 2) == (ad % 2) ? a[ad] : b[ad]);
				
				if(co < la) continue;
				dp[mk | (1 << ad)][co] = min(dp[mk][la] + iv, dp[mk | (1 << ad)][co]);
			}
		}
	}
	
	
	ll ans = Inf;
	for(int i = 0; i <= 50; i++) ans = min(ans, dp[(1 << n) - 1][i]);
	cout << (ans > n*n ? -1 : ans) << '\n';
	return 0;
}