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 <bits/stdc++.h>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n;
int a[51],b[51];
int dp[1<<18][51];
int INF=1<<25;

int main(void){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
	}
	for(int i=0;i<n;i++){
		scanf("%d",&b[i]);
	}
	for(int i=0;i<(1<<n);i++){
		for(int j=0;j<=50;j++){
			dp[i][j]=INF;
		}
	}
	dp[0][0]=0;
	for(int i=0;i<(1<<n);i++){
		int pos=__builtin_popcount(i);
		for(int j=0;j<=50;j++){
			if(dp[i][j]==INF)continue;
			int t_pos=0;
			//printf("%d %d %d %d\n",i,j,dp[i][j],pos);
			for(int k=0;k<n;k++){
				if((i>>k) & 1)continue;
				int flag=(k-pos+100)%2;
				//printf("k=%d flag=%d t_pos=%d\n",k,flag,t_pos);
				if(flag==0){
					if(j<=a[k]){
						dp[i+(1<<k)][a[k]]=min(dp[i+(1<<k)][a[k]],dp[i][j]+t_pos);
					}
				}else{
					if(j<=b[k]){
						dp[i+(1<<k)][b[k]]=min(dp[i+(1<<k)][b[k]],dp[i][j]+t_pos);
					}
				}
				t_pos++;
			}
		}
	}
	int ans=INF;
	for(int i=0;i<=50;i++){
		ans=min(ans,dp[(1<<n)-1][i]);
	}
	printf("%d\n",ans==INF?-1:ans);
	return 0;
}