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>
using namespace std;

typedef pair<int,int> pii;

const int N = 50;
const int INF = 1e9;
int n;
int li[4][N+5];

int cnt_bit(int x){
	int ret = 0;
	while(x > 0){
		ret += x&1;
		x/=2;
	}
	return ret;
}

bool cek_bit(int x, int mask){
	return (1<<x) & mask;
}

int tog_bit(int x, int mask){
	return (1<<x) ^ mask;
}

int memo[(1<<18)][52];

int dp(int id, int mask, int last){
	if(mask == (1<<n)-1) return 0;
	int &ret = memo[mask][last];
	if(ret!=-1) return ret;
	ret = INF;
	for(int i=0;i<n;i++){
		if(cek_bit(i,mask)) continue;
		int pake = -1;
		if(abs(i-id)&1){
			pake = li[1][i];
		} else {
			pake = li[0][i];
		}
		if(pake >= last){
			int co = 0;
			for(int j=0;j<i;j++){
				if(!cek_bit(j,mask)) co++;
			}
//			for(int j=0;j<n;j++) cout << (cek_bit(j,mask) ? 1 : 0);
//			cout << " >> " << i << " ke " << id << " cost " << co << " dengan " << pake << endl;
			ret = min(ret,dp(id+1,tog_bit(i,mask),pake) + co);
		}
	}
	return ret;
}


int main(){
	memset(memo,-1,sizeof(memo));
	scanf("%d",&n);
	for(int j=0;j<2;j++){
		for(int i=0;i<n;i++){
			scanf("%d",&li[j][i]);
		}
	}
	int ans = dp(0,0,0);
	if(ans == INF) ans = -1;
	printf("%d\n",ans);
	return 0;
}