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 <vector>
using namespace std;
int INF = 10000;
int main(){
	int N;
	cin >> N;
	vector<int> A(N);
	for (int i = 0; i < N; i++){
		cin >> A[i];
	}
	vector<int> B(N);
	for (int i = 0; i < N; i++){
		cin >> B[i];
	}
	vector<vector<int>> dp(1 << N, vector<int>(N, INF));
	for (int i = 0; i < N; i++){
		dp[1 << i][i] = i;
	}
	for (int i = 0; i < (1 << N); i++){
		for (int j = 0; j < N; j++){
			if (i >> j & 1){
				int num;
				if (j % 2 != __builtin_parity(i)){
					num = A[j];
				} else {
					num = B[j];
				}
				int cnt = 0;
				for (int k = 0; k < N; k++){
					if (!(i >> k & 1)){
						int num2;
						if (k % 2 == __builtin_parity(i)){
							num2 = A[k];
						} else {
							num2 = B[k];
						}
						if (num <= num2){
							dp[i | 1 << k][k] = min(dp[i | 1 << k][k], dp[i][j] + cnt);
						}
						cnt++;
					}
				}
			}
		}
	}
	int ans = INF;
	for (int i = 0; i < N; i++){
		ans = min(ans, dp[(1 << N) - 1][i]);
	}
	if (ans == INF){
		cout << -1 << endl;
	} else {
		cout << ans << endl;
	}
}