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 collections import defaultdict
from operator import itemgetter
N = int(input())
A = list(map(int,input().split()))
B = list(map(int,input().split()))

ans = 1000
for n in range(2**N):
    n = bin(n)[2:]
    if n.count('1')%2!=0:
        continue
    n = '0'*(N-len(n))+n
    l = []
    for i in range(N):
        if n[i]=='0':
            l.append([A[i],i])
        else:
            l.append([B[i],i+1])
    inversion = 0
    for i in range(N):
        if inversion>=ans:
            break
        for j in range(i+1,N):
            if l[i][0]>l[j][0]:
                inversion += 1
    if inversion>=ans:
        continue
    l.sort(key=itemgetter(0))
    d = defaultdict(list)
    for i in range(N):
        v,j = l[i]
        if (i-j)%2==1:
            d[v].append(i)
    possible = True
    for v in d:
        if len(d[v])%2==1:
            possible = False
        else:
            for i in range(len(d[v])//2):
                if (d[v][2*i+1]-d[v][2*i])%2==1:
                    inversion += d[v][2*i+1]-d[v][2*i]
                else:
                    possible = False
                    break
        if not possible:
            break
    if possible:
        ans = min(ans,inversion)

if ans == 1000:
    ans = -1
print(ans)