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>
using namespace std;

int N, A[19], B[19];
int dp[1 << 19][19];

int main() {
	cin >> N;
	for (int i = 0; i < N; i++) cin >> A[i];
	for (int i = 0; i < N; i++) cin >> B[i];

	for (int i = 0; i < (1 << N); i++) {
		for (int j = 0; j <= N; j++) dp[i][j] = (1 << 30);
	}

	dp[0][N] = 0;
	for (int i = 0; i < (1 << N); i++) {
		for (int j = 0; j <= N; j++) {
			if (dp[i][j] == (1 << 30)) continue;

			int bit[19], rui[19], cnts = 0;
			for (int k = 0; k < N; k++) { bit[k] = 0; if ((i & (1 << k)) != 0) bit[k] = 1; cnts += bit[k]; }
			for (int k = 0; k < N; k++) rui[k] = bit[k];
			for (int k = N - 2; k >= 0; k--) rui[k] += rui[k + 1];

			int preval = A[j];
			if (j == N) preval = -1000000000;
			else if (abs((cnts - j)) % 2 == 0) preval = B[j];

			// 次に何を取るか
			for (int k = 0; k < N; k++) {
				if (bit[k] == 1) continue;
				int r = A[k]; if (abs(cnts - k) % 2 == 1) r = B[k];
				if (preval <= r) {
					dp[i + (1 << k)][k] = min(dp[i + (1 << k)][k], dp[i][j] + rui[k]);
				}
			}
		}
	}
	
	int ans = (1 << 30);
	for (int i = 0; i < N; i++) ans = min(ans, dp[(1 << N) - 1][i]);

	if (ans == (1 << 30)) cout << "-1" << endl;
	else cout << ans << endl;
	return 0;
}