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

	int n;
	cin >> n;
	vector<int> a(n), b(n);
	for(int i=0; i<n; i++)
		cin >> a[i];
	for(int i=0; i<n; i++)
		cin >> b[i];

	int ans = -1;

	for(int r=0; r<(1<<n); r++) {

		int rev = 0;//反面的數量
		vector<int> v(n);
		for(int i=0; i<n; i++) {
			if(r&(1<<i)) v[i] = a[i];
			else v[i] = b[i], rev++;
		}
		if(rev&1) continue;

		vector<int> sorted(v);
		sort(sorted.begin(), sorted.end());

		vector<int> p(n, -1);
		for(int i=0; i<n; i++) {
			for(int j=0; j<n; j++) {
				if((i-j+2*n)%2 == ((r&(1<<j))?1:0)) continue;
				if(p[j]==-1 && v[j]==sorted[i]) {
					p[j] = i;
					break;
				}
			}
		}

		bool success = true;
		for(int i=0; i<n; i++) {
			if(p[i]==-1) {
				success = false;
				break;
			}
		}
		if(!success) continue;

		int cnt = 0;
		for(int i=0; i<n; i++) {
			for(int j=i+1; j<n; j++) {
				if(p[j]<p[i]) cnt++;
			}
		}
		ans = (ans==-1) ? cnt : min(ans, cnt);
	}
	cout << ans << "\n";
	return 0;
}