Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3
3 4 3
3 2 3

Sample Output 1

1

By doing the operation once with i = 1, we have a sequence [2, 3, 3] facing up, which is non-decreasing.

Sample Input 2

2
2 1
1 2

Sample Output 2

-1

After any number of operations, we have the sequence [2, 1] facing up, which is not non-decreasing.

Sample Input 3

4
1 2 3 4
5 6 7 8

Sample Output 3

0

No operation may be required.

Sample Input 4

5
28 15 22 43 31
20 22 43 33 32

Sample Output 4

-1

Sample Input 5

5
4 46 6 38 43
33 15 18 27 37

Sample Output 5

3

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

int main()
{
	int i, N, A[20], B[20];
	scanf("%d", &N);
	for (i = 1; i <= N; i++) scanf("%d", &(A[i]));
	for (i = 1; i <= N; i++) scanf("%d", &(B[i]));
	
	const int sup = 10000, bit[19] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144};
	int j, k, l, dp[262144][51] = {}, X[20][20], tmp, count, min;
	for (i = 1; i <= N; i++) {
		for (j = 1; j <= N; j++) X[i][j] = ((i + j) % 2 == 0)? A[i]: B[i];
	}
	for (k = 1; k < bit[N]; k++) {
		for (j = 0; j <= 50; j++) dp[k][j] = sup;
	}
	for (k = 1; k < bit[N]; k++) {
		for (i = 0, count = 0; bit[i] <= k; i++) if ((k | bit[i]) == k) count++;
		for (i = 0, l = 0; bit[i] <= k; i++) {
			if ((k | bit[i]) != k) continue;
			tmp = k ^ bit[i];
			if (dp[tmp][X[i+1][N-count+1]] < sup) {
				min = dp[tmp][X[i+1][N-count+1]] + abs(i - l - (N - count));
				for (j = 0; j <= X[i+1][N-count+1]; j++) if (dp[k][j] > min) dp[k][j] = min;
			}
			l++;
		}
	}
	
	if (dp[bit[N]-1][0] < sup) printf("%d\n", dp[bit[N]-1][0]);
	else printf("-1\n");
	fflush(stdout);
	return 0;
}