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 N 20
using namespace std;
int n;
int a[N],b[N];
int dp[1<<19][51];
int Abs(int a){
	return a > 0 ? a : -a;
}
int dfs(int hav, int ma,int cnt)
{
	if(cnt == n) return 0;
	if(dp[hav][ma] != -1) return dp[hav][ma];
	dp[hav][ma] = 10086;
	int dx,ans,tot = 0;
	for(int i = 0; i < n;i ++){
		if(hav & (1 << i)){
			tot ++;
			continue;
		}
		if(Abs(i - cnt) & 1) dx = b[i];
		else dx = a[i];
		if(dx < ma) continue;
		ans = dfs(hav | (1 << i), dx, cnt + 1);
		if(ans > 1000) continue;
		dp[hav][ma] = min(ans + cnt - tot, dp[hav][ma]);
	}
	return dp[hav][ma];
}
int main(){
	memset(dp, -1, sizeof(dp));
	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]);
	}
	int ans = dfs(0,0,0);
	if(ans > 1000) puts("-1");
	else printf("%d\n", ans);
	return 0;
}