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 <iostream>
#include <algorithm>
#include <utility>
#include <vector>
#define llint long long
#define inf 1e18

using namespace std;
typedef pair<llint, llint> P;

llint n;
llint a[18][2];
llint dp[1<<18][18][2];
llint pop[1<<18];

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n;
	for(int i = 0; i < n; i++) cin >> a[i][0];
	for(int i = 0; i < n; i++) cin >> a[i][1];
	
	llint N = 1<<n;
	for(int i = 1; i < N; i++) pop[i] = pop[i&(i-1)] + 1;
	
	for(int i = 0; i < N; i++){
		for(int j = 0; j < n; j++){
			for(int k = 0; k < 2; k++){
				dp[i][j][k] = inf;
			}
		}
	}
	dp[0][0][0] = 0;
	for(int i = 0; i < N; i++){
		for(int j = 0; j < n; j++){
			for(int k = 0; k < 2; k++){
				llint cnt = 0;
				for(int l = 0; l < n; l++){
					if(i & (1<<l)) continue;
					cnt++;
					llint dif = (pop[i]+l)%2, px = a[j][k], nx = a[l][dif];
					//cout << i << " " << j << 2 " << k << endl;
					if(i != 0 && nx < px) continue;
					dp[i|(1<<l)][l][dif] = min(dp[i|(1<<l)][l][dif], dp[i][j][k] + cnt-1);
				}
			}
		}
	}
	
	llint ans = inf;
	for(int j = 0; j < n; j++){
		for(int k = 0; k < 2; k++){
			ans = min(ans, dp[N-1][j][k]);
		}
	}
	if(ans > inf/2) cout << -1 << endl;
	else cout << ans << endl;
	
	return 0;
}