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 <iostream>
#include <cmath>
#include <cstring>

using namespace std;

typedef long long ll;
#define x first
#define y second
#define mp make_pair

const ll MAXN = 18;
ll n, a[MAXN + 5][2];
ll dp[(1 << 20) + 5][55];
const ll inf = (ll)1e9;

ll rek(int mask, int prev){
	if(!mask)
		return 0;
	ll &ret = dp[mask][prev];
	if(ret != -1)
		return ret;
	ret = inf;
	for(int i = 0 ; i < n ; ++i){
		if(!(mask & (1 << i)))
			continue;
		int cur = n - __builtin_popcount(mask);
		int br = cur - (i - __builtin_popcount(mask & ((1 << i) - 1)));
		//cout << mask << ' ' << cur << ' ' << br << ' ' << i << endl;
		cur = abs(cur - i);
		int sad = a[i][(cur & 1)];
		if(sad >= prev){
			ret = min(ret, rek(mask ^ (1 << i), sad) + br);
		}
	}
	return ret;
}

int main(){
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n;
for(int i = 0 ; i < n ; ++i){
	cin >> a[i][0];
}
for(int i = 0 ; i < n ; ++i){
	cin >> a[i][1];
}
memset(dp, -1, sizeof(dp));
ll sol = rek((1 << n) - 1, 0);
if(sol >= inf){
	cout << -1 << endl;
}
else{
	cout << sol << endl;
}

return 0;
}