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 <vector>
#include <cmath>
#include <algorithm>
#define rep(i, n) for(i = 0; i < n; i++)
using namespace std;

void chmin(int &a, int b) { a = min(a, b); }

int INF = 1e+9;
int n;
int a[18];
int b[18];
int dp[1 << 18][18];

int main() {
	int i, j, k;
	
	cin >> n;
	rep(i, n) cin >> a[i];
	rep(i, n) cin >> b[i];
	
	rep(i, (1 << n)) rep(j, n) dp[i][j] = INF;
	
	rep(i, n) {
		dp[1 << i][i] = 0;
	}
	
	rep(i, (1 << n) - 1) {
		int bcnt = 0;
		rep(j, n) if ((i >> j) % 2) bcnt++;
		
		rep(j, n) {
			if (dp[i][j] >= INF) continue;
			
			int last;
			if (bcnt % 2 != j % 2) { last = a[j]; }
			else { last = b[j]; }
			
			rep(k, n) {
				if ((i >> k) % 2) { continue; }
				
				int now;
				if ((bcnt + 1) % 2 != k % 2) { now = a[k]; }
				else { now = b[k]; }
				if (last > now) continue;

				int cst = 0;
				int l;
				for (l = k + 1; l < n; l++) if ((i >> l) & 1) cst++;
				
				chmin(dp[i + (1 << k)][k], dp[i][j] + cst);
			}
		}
	}
	
	int ans = INF;
	rep(j, n) {
		chmin(ans, dp[(1 << n) - 1][j]);
	}
	if (ans >= INF) cout << -1 << endl;
	else cout << ans << endl;
	return 0;
}