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, l, r) for (register int i = l; i <= r; i++)
#define per(i, r, l) for (register int i = r; i >= l; i--)
#define srep(i, l, r) for (register int i = l; i < r; i++)
#define sper(i, r, l) for (register int i = r; i > l; i--)
#define maxn 20
#define ll long long int 
using namespace std;

const int inf = 2000000020;

int n;
int v[maxn][2];

bool vis[1 << maxn][maxn];
int dp[1 << maxn][maxn];
int DP(int op, int last){
	if (op == ((1 << n) - 1)) return 0;
	if (vis[op][last]) return dp[op][last];
	vis[op][last] = 1;
	int& ans = dp[op][last];
	ans = inf;
	int now = 0, sum = 0;
	rep(i, 1, n) if (op & (1 << (i - 1))) sum++;
	per(i, n, 1){
		if (op & (1 << (i - 1))) {now++; continue;}
		if (v[i][((sum + 1) & 1) ^ (i & 1)] < v[last][(sum & 1) ^ (last & 1)]) continue;
		ans = min(ans, DP(op | (1 << (i - 1)), i) + now);
	} 
	return ans;
} 
int main(){
	
		//freopen("D.in", "r", stdin);
	
	cin >> n;
	rep(i, 1, n) cin >> v[i][0];
	rep(i, 1, n) cin >> v[i][1];
	
	int ans = DP(0, 0);
	if (ans == inf) printf("-1");
	else printf("%d", ans);
	return 0;
}