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>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) begin(v),end(v)
#define fi first
#define se second
template<typename A, typename B> inline bool chmax(A &a, B b) { if (a<b) { a=b; return 1; } return 0; }
template<typename A, typename B> inline bool chmin(A &a, B b) { if (a>b) { a=b; return 1; } return 0; }
using ll = long long;
using pii = pair<int, int>;
constexpr ll INF = 1ll<<30;
constexpr ll longINF = 1ll<<60;
constexpr ll MOD = 1000000007;
constexpr bool debug = 0;
//---------------------------------//

int N;
int A[18], B[18];

int dp[1<<18][51]; // 残り集合 最後の値がk

int main() {
	cin >> N;
	REP(i, N) cin >> A[i];
	REP(i, N) cin >> B[i];
	
	fill(dp[0], dp[1<<18], INF);
	
	dp[(1<<N)-1][0] = 0;
	
	for (int s = (1<<N)-1; s >= 0; --s) {
		REP(k, 51) {
			if (dp[s][k] == INF) continue;
			int cnt = 0;
			REP(j, N) if (s >> j & 1 ^ 1) ++cnt;
			// cnt番目に置く
			
			int c = 0;
			REP(j, N) {
				if (s >> j & 1) {
					int dif = abs(cnt - j);
					int cur = dif & 1 ? B[j] : A[j];
					if (cur >= k) chmin(dp[s^(1<<j)][cur], dp[s][k] + c);
					++c;
				}
			}
		}
	}
	
	int ans = INF;
	REP(i, 51) chmin(ans, dp[0][i]);
	if (ans == INF) ans = -1;
	cout << ans << endl;
	
	return 0;
}