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
from itertools import combinations
n = int(input())
As = list(map(int, input().split()))
Bs = list(map(int, input().split()))

even = []   #偶数番目で表になる
odd = []    #奇数番目で表になる
for i,(a,b) in enumerate(zip(As,Bs)):
    if i%2 == 0:
        even.append(a)
        odd.append(b)
    else:
        even.append(b)
        odd.append(a)



def check(n, even, odd):
    prev = -10**10
    ps = []
    for i in range(n):
        if i%2 == 0:
            if prev > even[i//2][0]:
                return False,-1
            else:
                ps.append(even[i//2][1])
            prev = even[i//2][0]
        else:
            if prev > odd[i//2][0]:
                return  False, -1
            else:
                ps.append(odd[i//2][1])
            prev = odd[i//2][0]
    return True, ps
    


def inversion(a):
    n = len(a)
    cnt = 0
    for i in range(n):
        for j in range(i+1,n):
            if a[i] > a[j]:
                cnt +=1
    return cnt
 
#偶数としてpickupするやつ
INF = 10**10
ans = 10**10
for l in combinations(range(n), (n-1)//2+1):
    cnt = 0
    even_selected = [(even[i],i) for i in range(n) if i in l]
    odd_selected = [(odd[i],i) for i in range(n) if i not in l]
    # こいつらを昇順に並べられるか
    even_selected.sort()
    odd_selected.sort()
    is_possible, a = check(n, even_selected, odd_selected)
    # 並び順が正しければ
    if is_possible:
        cnt = inversion(a)
        ans = min(cnt, ans)
        
if ans == INF:
    print(-1)
else:
    print(ans)        