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>
#define va first
#define vb second
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

int N;
int dp[18][1<<18];
pii A[20];

int main() {
	cin >> N;
	for (int i=0; i<N; i++) cin >> A[i].va;
	for (int i=0; i<N; i++) cin >> A[i].vb;

	for (int i=0; i<(1<<N); i++) {
		for (int j=0; j<N; j++) dp[j][i] = 2e9;
	}
	for (int i=0; i<N; i++) dp[i][1<<i] = 0;

	for (int i=0; i<(1<<N); i++) {
		for (int j=0; j<N; j++) {
			if (!(i>>j&1)) continue;
			int prv = i ^ (1<<j);
			int idx = 0, cost = 0, num;
			for (int k=0; k<N; k++) {
				idx += (prv>>k&1);
				if (k>j) cost += (prv>>k&1);
			}
			if (idx%2 == j%2) num = A[j].va;
			else num = A[j].vb;

			for (int k=0; k<N; k++) {
				if (!(prv>>k&1)) continue;
				int t;
				if (k%2 == idx%2) t = A[k].vb;
				else t = A[k].va;
				if (t > num) continue;
				dp[j][i] = min(dp[j][i], dp[k][prv]+cost);
			}
		}
	}

	int ans = 2e9;
	for (int i=0; i<N; i++)
		ans = min(ans, dp[i][(1<<N)-1]);
	cout << (ans > 1e9 ? -1 : ans);
}