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 rep(i, a, n) for(int i = a; i < n; i++)
#define REP(i, n) rep(i, 0, n)
#define int long long
using namespace std;
typedef pair<int, int> P;
const int mod = 1000000007;
const int INF = 1e9;

int dp[1 << 18][18];

signed main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	vector<int> a(n), b(n);
	rep(i, 0, n) cin >> a[i];
	rep(i, 0, n) cin >> b[i];
	rep(i, 0, 1 << n) rep(j, 0, n) dp[i][j] = INF;
	rep(i, 0, n) dp[1 << i][i] = 0;
	for(int mask = 0; mask < (1 << n); mask++){
		int c = __builtin_popcountll(mask);
		for(int i = 0; i < n; i++){
			if(dp[mask][i] == INF) continue;
			if(mask & (1 << i)){
				int r = 0;
				for(int j = 0; j < n; j++){
					if(mask & (1 << j)){
						r++;
						continue;
					}
					int pre = (((c - 1) % 2 == i % 2) ? a[i] : b[i]);
					int nxt = ((c % 2 == j % 2) ? a[j] : b[j]);
					if(pre <= nxt){
						dp[mask | (1 << j)][j] = min(dp[mask | (1 << j)][j], dp[mask][i] + abs(c - r));
					}
				}
			}
		}
	}
	int ans = INF;
	rep(i, 0, n) ans = min(ans, dp[(1 << n) - 1][i]);
	if(ans >= INF) ans = -1;
	cout << ans << endl;
}