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

N = int(input())

A = list(map(int,input().split()))
B = list(map(int,input().split()))
ans = float("inf")

for ii in range(2 ** N):

    s = format(ii,'b').zfill(N)
    on = 0
    
    for j in s:
        if j == "1":
            on += 1

    if on != (N+1)//2:
        continue

    q = []
    q2 = []

    for i in range(N):
        if s[i] == "1":
            if i % 2 == 0:
                q.append([A[i],i])
            else:
                q.append([B[i],i])
        else:
            if i % 2 != 0:
                q2.append([A[i],i])
            else:
                q2.append([B[i],i])           

    q.sort()
    q2.sort()
    q2.append([float("inf"),0])
    #print (q,q2)

    flag = True
    nmax = 0

    for i in range((N+1)//2):


        if nmax > q[i][0]:
            flag = False
        nmax = q[i][0]

        if nmax > q2[i][0]:
            flag = False
        nmax = q2[i][0]

    if flag:
        r = []
        for i in range((N+1)//2):
            r.append(q[i][1])
            r.append(q2[i][1])

        #print (r)

        now = 0
        z = [i for i in range(N)]
        for i in range(N):
            for j in range(len(z)):
                if z[j] == r[i]:
                    now += j
                    del z[j]
                    break
            #print (z,now)
        ans = min(now,ans)

if ans == float("inf"):
    ans = -1
print (ans)
