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;

const int N = 1 << 18;
const int INF = 1e6;

int n;
int a[N + 5], b[N + 5];
int dp[N + 5][51];
int val[N + 5];

int f(int id, int mask){
	int ret = 0;
	for(int i=0; i<id; i++){
		if((1 << i) & mask) continue;
		ret++;
	}
	return ret;
}

int solve(int mask, int last, int vlast){
	if(mask == (1 << n) - 1) return 0;
	int& ret = dp[mask][vlast];
	if(ret != -1) return ret;
	ret = INF;
	for(int i=0; i<n; i++){
		if(mask & (1 << i)) continue;
		int v = abs(val[mask] - i) % 2 == 0 ? a[i] : b[i];
		if(vlast > v) continue;
		ret = min(ret, solve(mask | (1 << i), i, v) + f(i,mask));
	}
	return ret;
}

int main(){
	ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
	for(int i=0; i<N; i++){
		int cnt = 0;
		for(int j=0; j<18; j++) if(i & (1 << j)) cnt++;
		val[i] = cnt;
	}
	cin >> n;
	for(int i=0; i<n; i++) cin >> a[i];
	for(int i=0; i<n; i++) cin >> b[i];
	memset(dp,-1,sizeof dp);
	int ans = INF;
	for(int i=0; i<n; i++){
		ans = min(ans, solve(1 << i, i, (i % 2 == 0) ? a[i] : b[i]) + i);
	}
	if(ans >= 1000) ans = -1;
	cout << ans << endl;
	return 0;
}