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 mp make_pair
#define pb push_back
#define fi first
#define se second

const int N = 19, inf = 1e8;

int n, a[N], b[N], dp[1 << N][51];

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0);
	cin >> n;
	for(int i = 1; i <= n; i++) cin >> a[i];
	for(int i = 1; i <= n; i++) cin >> b[i];
	memset(dp, 62, sizeof(dp));
	dp[0][1] = 0;
	for(int mask = 0; mask < (1 << n) - 1; mask++){
		int p = __builtin_popcount(mask) + 1;
		for(int val = 1; val <= 50; val++){
			if(dp[mask][val] >= inf) continue;
			int cost = 0;
			for(int i = 1; i <= n; i++){
				if(mask & (1 << (i - 1))) continue;
				int newmask = mask ^ (1 << (i - 1));
                int diff = abs(p - i) % 2, cur = a[i];
                if(diff == 1) cur = b[i];
                if(cur < val){
					cost++;
					continue;
                }
                dp[newmask][cur] = min(dp[newmask][cur], dp[mask][val] + cost);
				cost++;
			}
		}
	}
	int ans = inf;
	for(int i = 1; i <= 50; i++){
		ans = min(ans, dp[(1 << n) - 1][i]);
	}
	if(ans >= inf) cout << -1;
	else cout << ans;
}