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>
#include<iostream>
using namespace std;

const int MAXN = 18;
int n,a[MAXN],b[MAXN],dp[1<<MAXN][MAXN];

int main(){
	cin >> n;
	for(int i=0;i<n;i++){
		cin >> a[i];
	}
	for(int i=0;i<n;i++){
		cin >> b[i];
	}
	for(int i=0;i<n;i++){
		for(int j=0;j<(1<<n);j++){
			dp[j][i]=1e9;
		}
	}
	for(int i=0;i<n;i++)dp[1<<i][i]=0;
	// dp[0][0]=0;
	for(int mask=0;mask<(1<<n);mask++){
		for(int i=0;i<n;i++){
			int p = __builtin_popcount(mask);
			int x;
			if(i%2==(p-1)%2)x = a[i];
			else x = b[i];
			for(int j=0;j<n;j++){
				if((mask>>j)&1)continue;
				int y = (j%2==p%2)?a[j]:b[j];
				if(y<x)continue;
				int cost = dp[mask][i]+__builtin_popcount(mask>>j);
				dp[mask|(1<<j)][j] = min(dp[mask|(1<<j)][j],cost);
			}
		}
	}
	int ans = 1e9;
	for(int i=0;i<n;i++)ans = min(ans,dp[(1<<n)-1][i]);
	if(ans>=1e9)ans = -1;
	cout << ans << endl;

	return 0;
}