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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int inf = 1e9;

int popcount(int n){
    n = (n & 0x55555555) + (n >> 1 & 0x55555555);
    n = (n & 0x33333333) + (n >> 2 & 0x33333333);
    n = (n & 0x0f0f0f0f) + (n >> 4 & 0x0f0f0f0f);
    n = (n & 0x00ff00ff) + (n >> 8 & 0x00ff00ff);
    return (n & 0x0000ffff) + (n >>16 & 0x0000ffff);
}

int main(){
	int n;
	cin >> n;
	vector<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];
	}

	// dp[i][j]: iの組み合わせで選んで、最後がjのときの最小コスト
	vector<vector<int>> dp(1<<n, vector<int>(n, inf));
	for(int i=0; i<n; i++){
		dp[1<<i][i] = i;
	}
	for(int i=1; i<(1<<n); i++){
		int count = popcount(i);
		for(int j=0; j<n; j++){
			if(!(i>>j&1)) continue;
			if(dp[i][j] == inf) continue;
			//cout << i << " " << j << endl;
			int last = ((count-1-j)%2==0)? a[j]: b[j];
			for(int k=0; k<n; k++){
				if(i>>k&1) continue;
				//cout << i << " " << j << " " << k << endl;
				int newlast = ((count-k)%2==0)? a[k]: b[k];
				//cout << "last " << last << " " << newlast << endl;
				if(newlast < last) continue;
				int cost = max(k-count, 0);
				int newset = i|(1<<k);
				dp[newset][k] = min(dp[newset][k], dp[i][j]+cost);
			}
		}
	}
	/*
	for(int i=0; i<(1<<n); i++){
		cout << "i= " << i << endl;
		for(int j=0; j<n; j++){
			cout << dp[i][j] << endl;
		}
	}
	*/
	int ans = *min_element(dp[(1<<n)-1].begin(), dp[(1<<n)-1].end());
	if(ans == inf) ans = -1;
	cout << ans << endl;
	return 0;
}