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;
#define int long long
const int MOD = 1000000007;

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<int> A(N);
	vector<int> B(N);
	int INF = (int)1 << 60;
	int res = INF;
	for (int i = 0; i < N; i++) {
		cin >> A[i];
	}
	for (int i = 0; i < N; i++) {
		cin >> B[i];
	}

	for (int i = 0; i < (1 << N); i++) {
		int ans = 0;
		vector<int> V(N, 0);
		bool ok = true;
		for (int p = 0; p < N; p++) {
			int mnj = -1;
			int mn = INF;
			int tmn = INF;
			int ad = -1;
			int tmp = 0;
			for (int j = 0; j < N; j++) {
				if (V[j] == 0) {
					if (i & (1 << j)) {
						tmn = min(tmn, A[j]);
						if (mn > A[j]) {
							if (abs(p - j) % 2 == 0) {
								mn = A[j];
								mnj = j;
								ad = tmp;
							}
						}
					} else {
						tmn = min(tmn, B[j]);
						if (abs(p - j) % 2 == 1) {
							if (mn > B[j]) {
								mn = B[j];
								mnj = j;
								ad = tmp;
							}
						}
					}
					tmp++;
				}
			}
			if (mn != tmn || mnj == -1) {
				ok = false;
				break;
			}
			ans += ad;
			V[mnj] = 1;
		}
		if (ok) {
			res = min(res, ans);
		}
	}
	if (res == INF) res = -1;
	cout << res << endl;
}