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 pb push_back
#define mp make_pair
#define ll long long
#define x first
#define y second
#define pii pair<int , int>
#define sz(a) (int)(a.size())
const int INF = 1000 * 1000 * 1000;
const int mod = 998244353;
const int MAX = 1000 * 1000 + 47;
using namespace std;
int a[2][20];
int dp[1 << 18][18];
int main()
{
	ios_base::sync_with_stdio(0);
	int n;
	cin >> n;
	for(int i = 0; i < 2; i++)
		for(int j = 0; j < n; j++)
			cin >> a[i][j];
	for(int i = 0; i < (1 << n); i++)
		for(int j = 0; j < n; j++)
			dp[i][j] = INF;
	for(int i = 0; i < n; i++)
		dp[(1 << i)][i] = 0;
	for(int i = 1; i < (1 << n); i++)
		for(int j = 0; j < n; j++)
			for(int k = 0; k < n; k++)
			{
				if((i >> k) & 1)
					continue;
				int cnt = __builtin_popcount(i);
				int val = a[(j + cnt - 1) & 1][j];
				int val2 = a[(k + cnt) & 1][k];
				if(val <= val2)
					dp[i | (1 << k)][k] = min(dp[i | (1 << k)][k] , dp[i][j] + __builtin_popcount(i >> k));
			}
	int ans = INF;
	for(int i = 0; i < n; i++)
	{
		ans = min(ans , dp[(1 << n) - 1][i]);
	}
	cout << (ans == INF ? -1 : ans);
	
}