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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
int d[1<<18][51][2];
int MIN(int a,int b){return a<b?a:b;}
int main(){
  int ans=1e9,e,c,i,j,l,f,k,n,a[20],b[20];
  for(i=0;i<(1<<18);i++){
    for(j=0;j<51;j++){
      for(k=0;k<2;k++)d[i][j][k]=1e9;
    }
  }
  d[0][0][0]=0;
  scanf("%d",&n);
  for(i=0;i<n;i++)scanf("%d",&a[i]);
  for(i=0;i<n;i++)scanf("%d",&b[i]);
  e=1<<n;
  for(i=0;i<e;i++){
    for(j=0;j<51;j++){
      for(k=0;k<n;k++){
	if(i&(1<<k))continue;
	c=0;
	for(l=i;l;l/=2)c+=l%2;
	for(f=0;f<2;f++){
	  if(c%2==k%2){
	    if(j>a[k])continue;
	    if(c>k)d[i|(1<<k)][a[k]][f]=MIN(d[i|(1<<k)][a[k]][f],d[i][j][f]+c-k);
	    else   d[i|(1<<k)][a[k]][f]=MIN(d[i|(1<<k)][a[k]][f],d[i][j][f]);
	  }
	  else{
	    if(j>b[k])continue;
	    if(c>k)d[i|(1<<k)][b[k]][1-f]=MIN(d[i|(1<<k)][b[k]][1-f],d[i][j][f]+c-k);
	    else   d[i|(1<<k)][b[k]][1-f]=MIN(d[i|(1<<k)][b[k]][1-f],d[i][j][f]);
	  }
	}
      }
    }
  }
  /*for(i=0;i<e;i++){
    for(j=0;j<51;j++){
      for(f=0;f<2;f++)printf("%d ",d[i][j][f]);
    }printf("\n");
    printf("\n");
    }//*/
  for(i=1;i<51;i++)ans=MIN(ans,d[e-1][i][0]);
  printf("%d\n",ans<1e8?ans:-1);
  return 0;
}