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 rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef pair<int,int>P;
typedef long long ll;

const int MOD=1000000007;
const int INF=0x3f3f3f3f;
const ll INFL=0x3f3f3f3f3f3f3f3f;

int cal[18][1<<18];
int dp[1<<18][50];

int a[20],b[20];

int main(){
	int n;cin>>n;
	rep(i,n)scanf("%d",&a[i]),a[i]--;
	rep(i,n)scanf("%d",&b[i]),b[i]--;
	rep(i,n)rep(j,1<<n){
		rep(k,n){
			if(k>i&&(j>>k&1))cal[i][j]++;
			if(k<i&&!(j>>k&1))cal[i][j]++;
		}
	}
	memset(dp,0x3f,sizeof(dp));
	dp[0][0]=0;
	rep(i,1<<n)rep(j,50){
		if(dp[i][j]==INF)continue;
		rep(k,n){
			if(i>>k&1)continue;
			int o=cal[k][i]%2;
			int D=(o==0?a[k]:b[k]);
			if(j>D)continue;
			dp[i|1<<k][D]=min(dp[i|1<<k][D],dp[i][j]+cal[k][i]);
		}
	}
	int Min=INF;
	rep(i,50){
		Min=min(Min,dp[(1<<n)-1][i]);
	}
	if(Min==INF)puts("-1");
	else cout<<Min/2<<endl;
}