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 debug(x) cerr<<#x<<'='<<x<<endl
#define set0(x) memset(x,0,sizeof(x))
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
template<typename T> void read(T &x){
	x = 0;char ch = getchar();ll f = 1;
	while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
	while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}

int dp[270000][160] = {0};

int n,a[20],b[20];
int main() {
	read(n);
	for(int i=0;i<n;i++)read(a[i]);
	for(int i=0;i<n;i++)read(b[i]);
	memset(dp,7,sizeof(dp));
	dp[(1<<n)-1][0] = 0;
	for(int i = (1<<n)-1;i;i--){
		int cto = n-__builtin_popcount(i);
		for(int j=0;j<160;j++){
			if(dp[i][j]>160)continue;
			//cout<<i<<' '<<j<<' '<<dp[i][j]<<endl;
			int fr = 0;
			for(int k=0;k<n;k++){
				if((i>>k)&1){
					int aft = cto-fr,mfr = k-fr;
					int cv = ((aft+mfr)%2 == 0)?a[k]:b[k];
					if(cv>=dp[i][j]){
						dp[i^(1<<k)][j+mfr] = min(dp[i^(1<<k)][j+mfr],cv);
					}
				}else{
					fr+=1;
				}
			}
		}
	}
	int ans = -1;
	for(int j=0;j<160;j++){
		if(dp[0][j]<100){
			ans = j;
			break;
		}
	}
	cout<<ans<<endl;
	return 0;
}