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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;
using ll = long long;
const ll INF = (int)1e9;

const int N = 18;
const int V = 50;
int dp[1 << N][V+1];
int as[N];
int bs[N];
int off[N][N];

void cap(int& v, int off) {
	if (v > off) v = off;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	int n;
	cin >> n;
	for (int i = 0; i < n; ++i) cin >> as[i];
	for (int i = 0; i < n; ++i) cin >> bs[i];
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) {
			off[i][j] = (i % 2 == j % 2) ? as[j] : bs[j];
		}
	}

	int h = 1 << n;
	for (int m = 0; m < h; ++m) {
		for (int j = 0; j <= V; ++j) dp[m][j] = INF;
	}
	dp[0][0] = 0;
	
	for (int m = 0; m < h; ++m) {
		int i = __builtin_popcount(m);
		for (int v = 0; v <= V; ++v) {
			int d = 0;
			for (int j = 0; j < n; ++j) {
				if (m & (1 << j)) continue;
				if (off[i][j] >= v) cap(dp[m | (1 << j)][off[i][j]], dp[m][v] + d);
				++d;
			}
		}
	}

	int res = INF;
	for (int v = 0; v <= V; ++v) res = min(res, dp[h-1][v]);
	cout << (res >= INF ? -1 : res) << '\n';
}