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;

int dp[1 << 18][18] , N , A[18] , B[18];

int main(){
	cin >> N; for(int i = 0 ; i < N ; ++i) cin >> A[i];
	for(int i = 0 ; i < N ; ++i) cin >> B[i];
	for(int i = 0 ; i < N ; ++i) if(i & 1) swap(A[i] , B[i]);
	memset(dp , 0x3f , sizeof(dp));
	for(int i = 0 ; i < N ; ++i) dp[1 << i][i] = i;
	for(int i = 1 ; i + 1 < 1 << N ; ++i)
		for(int j = 0 ; j < N ; ++j)
			if(dp[i][j] <= 1e4){
				int cnt = __builtin_popcount(i) , num = 0;
				for(int k = 0 ; k < N ; ++k)
					if(!(i >> k & 1)){
						if((cnt & 1) && A[j] > B[k] || !(cnt & 1) && B[j] > A[k]) continue;
						dp[i | (1 << k)][k] = min(dp[i | (1 << k)][k] , dp[i][j] + k - num);
					}else ++num;
			}
	int mn = 1e9; for(int i = 0 ; i < N ; ++i) mn = min(mn , dp[(1 << N) - 1][i]);
	printf("%d\n" , mn > 1e4 ? -1 : mn);
}