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;
typedef long long ll;

int N;
const int MAXN = 20;
const int big = MAXN*MAXN + 1;
int A[MAXN][2];
int dp[(1 << MAXN)][51][2];

int solve(int mask, int last, bool odd) {
	if (mask == 0) {
		return 0;
	}
	if (dp[mask][last][odd] >= 0) {
		return dp[mask][last][odd]; 
	}
	int best = big;	
	int count = 0;
	for (int i = 0; i < N; i++) {
		if (mask & (1 << i)) {
			int t = ((odd + i) % 2 ? A[i][1] : A[i][0]);
			if (t >= last) {
				int val = count + solve(mask ^ (1 << i), t, !odd);
				best = min(val, best);
			}
			count++;
		}
	}
	dp[mask][last][odd] = best;
	return best;
}

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);	
	cin >> N;
	for (int i=0; i < N; i++) {
		cin >> A[i][0];
	}
	int mask = (1 << N) - 1;
	for (int i=0; i < N; i++) {
		cin >> A[i][1];
	}
	memset(dp, -1, sizeof dp);
	int swaps = solve(mask, 0, 0);
	if (swaps >= big) {
		cout << -1 << "\n";
	} else {
		cout << swaps << "\n";
	}
}