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 loop in range(2 ** N):

    s = format(loop,'b').zfill(N) #sはどのカードが奇数番目かを表す。ビットの組み合わせを2^N通り入れて全探索。
    on = 0
    
    for j in s:
        if j == "1":
            on += 1

    if on != (N+1)//2: #sの立っているビットの数が奇数番目の位置の個数に合っていなかったら飛ばす
        continue

    q = []  #奇数番目に置くカードの集合
    q2 = [] #偶数番目に置くカードの集合

    for i in range(N): #カードの位置の偶奇に合わせて表裏を決める
        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]) #偶数番目は1枚少ないことがあるので番兵
    

    flag = True #昇順になっているかのフラグ
    nmax = 0    #現在の最大値(昇順判定に使う)

    for i in range((N+1)//2): #昇順判定プロセス


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

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

    if flag: #昇順だったらrにカードの番号(not 書かれている数字)を入れてく
        r = []
        for i in range((N+1)//2):
            r.append(q[i][1])
            r.append(q2[i][1])

        #ここから転倒数の計算(無理やりシミュレーション)
        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
        ans = min(now,ans) #求める答えは転倒数の最小値

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