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;
#define rep(i,n) for(int i=0; i<(n); i++)
int n,a[2][20],ans=1e9;
int main(){
	scanf("%d", &n);
	rep(o,2) rep(i,n) scanf("%d", a[o]+i);
	rep(bt,1<<n){
		int rev[40] = {}, c[] = {-2,-1};
		bool ok = true;
		vector<int> v[51];
		rep(i,n) v[a[bt>>i&1][i]].push_back(i);
		rep(i,51){
			for(int t: v[i]) rev[c[abs(c[1]-t)%2 == (bt>>t)%2]+=2] = t+1;
			if(abs(c[0]-c[1])>1) ok = false;
		}
		if(ok){
			constexpr int sz = 32;
			int sm = 0;
			vector<int> dat(sz+1);
			rep(i,n){
				for(int j=rev[i]; j; j-=j&-j) sm+=dat[j];
				for(int j=rev[i]; j<=sz; j+=j&-j) dat[j]++;
			}
			sm = n*(n-1)/2 - sm;
			ans = min(ans,sm);
		}
	}
	printf("%d\n",ans == (int)1e9 ? -1 : ans);
}