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>
typedef long long ll;
using namespace std;
const ll mod = 1000000007;
const ll mod998 = 998244353;
const ll intmax = 2147483647;
const ll llmax = 9223372036854775807;
const char sp = ' ';

struct a {
	int num;
	bool goal, now;
};

a t[18];
int N, A[18][2], res, v[18];
bool b[18];


int main() {
	cin >> N;
	for (int i = 0; i < N; i++) {
		cin >> A[i][1];
	}
	for (int i = 0; i < N; i++) {
		cin >> A[i][0];
	}
	res = 0xE869120;
	for (int i = 0; i < (1 << N); i++) {
		int c = 0;
		for (int j = 0; j < N; j++) {
			c += (i >> j) & 1;
			t[j].goal = (i >> j) & 1;
			t[j].now = 1;
			v[j] = t[j].num = A[j][(i >> j) & 1];
			b[j] = false;
		}
		if ((c & 1) != (N & 1))continue;
		sort(v, v + N);
		int cnt = 0;
		bool f = false;
		for (int j = 0; j < N; j++) {
			for (int k = j; k <= N; k++) {
				if (k == N) {
					f = true;
					break;
				}
				if (t[k].num == v[j] && (((k - j) & 1) ^ t[k].now^t[k].goal^1)) {
					for (int l = k - 1; l >= j; l--) {
						t[l + 1].num = t[l].num;
						t[l + 1].goal = t[l].goal;
						t[l + 1].now = (t[l].now ^ 1);
						cnt++;
					}
					break;
				}
			}
			if (f)break;
		}
		if (f)continue;
		res = min(cnt, res);
	}
	if (res == 0xE869120)res = -1;
	cout << res << endl;
}