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 a[19], b[19];
int f[300001][19];
int n;

int main() 
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	cin >> n;
	for (int i=1;i<=n;i++)
		cin >> a[i];
	for (int i=1;i<=n;i++)
		cin >> b[i];
	for (int i=0;i<=300000;i++)
		for (int j=0;j<=18;j++)
			f[i][j] = 1e9;
	for (int i=1;i<=n;i++)
		f[(1<<(i-1))][i] = 0;
	for (int mask = 1;mask < (1<<n);mask++)
	{
		int pos = __builtin_popcount(mask) + 1;
		for (int i=0;i<=n-1;i++)
			if ((1<<i) & mask)
			{
				int cnt = 0;
				for (int j=0;j<=n-1;j++)
				{
					if (!((1<<j) & mask))
					{
						int cost = pos - 1 - cnt;
						int pre_val = ((abs((i + 1) - (pos - 1)) % 2) == 0) ? a[i+1] : b[i+1];
						int cur_val = ((abs((j + 1) - pos) % 2) == 0) ? a[j+1] : b[j+1];
						if (cur_val >= pre_val)
							f[mask | (1<<j)][j+1] = min(f[mask | (1<<j)][j+1],f[mask][i+1] + cost);
					}
					else
						cnt++;
				}
			}
	}
	int res = 1e9;
	for (int i=1;i<=n;i++)
		res = min(res,f[(1<<n)-1][i]);
	if (res == 1e9) cout << -1;
	else cout << res;
	return 0;
}