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()))
lst=[]
for i in range(2**N):
    h=[[],[]]
    for j in range(N):
        if (i>>j)&1==1:
            h[(j+1)%2].append([B[j],j])
        else:
            h[j%2].append([A[j],j])
    if len(h[1])==N//2:
        lst.append(h)
lsls=[]
for h in lst:
    h[0].sort()
    h[1].sort()
    cnt=h[0][0][1]
    for i in range(N-1):
        if h[i%2][i//2][0]>h[(i+1)%2][(i+1)//2][0]:
            break
    else:
        hhh=[]
        for i in range(N):
            hhh.append(h[i%2][i//2][1])
        lsls.append(hhh)
inf=float("inf")
ans=inf
if lsls:
    for h in lsls:
        flag=[0]*N
        cnt=0
        for u in h:
            flag[u]=1
            cnt+=sum(flag[u+1:])
        ans=min(ans,cnt)
if ans==inf:
    print(-1)
else:
    print(ans)