Problem: 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.

Examples

Input

3
3 4 3
3 2 3


Output

1


Input

2
2 1
1 2


Output

-1


Input

4
1 2 3 4
5 6 7 8


Output

0


Input

5
28 15 22 43 31
20 22 43 33 32


Output

-1


Input

5
4 46 6 38 43
33 15 18 27 37


Output

3

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;

int main() {
	long long N, INF = 1000000000, ANS = INF;
	cin >> N;
	vector<vector<long long> > A(N, vector<long long>(2));
	for (int i = 0; i < N; i++) cin >> A[i][0];
	for (int i = 0; i < N; i++) cin >> A[i][1];
	vector<long long> V;
	for (int i = 0; i < N; i++) V.push_back(i % 2);
	sort(V.begin(), V.end());
	long long Even = (N + 1) / 2, Odd = N / 2;
	do {
		vector<pair<long long, long long> > E, O;
		for (int i = 0; i < N; i++) {
			if (V[i] == 0) E.push_back({ A[i][i % 2], i });
			else O.push_back({ A[i][(i + 1) % 2], i });
		}
		sort(E.begin(), E.end()), sort(O.begin(), O.end());
		bool check = true;
		vector<pair<long long, long long> > X;
		for (int i = 0; i < N; i++) {
			if (i % 2 == 0) X.push_back(E[i / 2]);
			else X.push_back(O[i / 2]);
		}
		for (int i = 0; i < N - 1; i++) {
			if (X[i].first > X[i + 1].first) check = false;
		}
		if (!check) continue;
		long long COUNT = 0;
		for (int i = 0; i < N; i++) {
			for (int j = i + 1; j < N; j++) {
				if (X[i].second > X[j].second) COUNT++;
			}
		}
		ANS = min(ANS, COUNT);
	} while (next_permutation(V.begin(), V.end()));
	cout << (ANS == INF ? -1 : ANS) << endl;
}