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()))

#0-indexed

#カードiが左からjの位置にあるときの表の数字
def getNum(i,j):
    return B[i] if abs(i-j)&1 else A[i]

#dp[使用済みカードを表すbit列][いちばん右のカードNo]=最小操作回数
INF=10**9
dp=[[INF for _ in range(N)] for __ in range(1<<N)]
for i in range(N):
    dp[1<<i][i] = i

for m in range(1,1<<N): #配置済みカードパターン
    cnt=bin(m).count('1') #配置済みカード数
    idx=dict() #未配置カードの位置
    tmp=cnt
    for i in range(N):
        if m & (1<<i) == 0:
            idx[i]=tmp
            tmp+=1
    for j in range(N): #新たに並べるカード
        if m & (1<<j):
            #もう使ったやつです
            continue
        num_j=getNum(j,cnt)
        for i in range(N): #配置済みのいちばん右にあるカード
            if m & (1<<i) == 0:
                #まだ配置済みじゃないです
                continue
            num_i=getNum(i,cnt-1)
            if num_i>num_j:
                #単調増加になりません
                continue
            dp[m|(1<<j)][j] = min(dp[m|(1<<j)][j], dp[m][i] + idx[j]-cnt)

ans=INF
for i in range(N):
    ans=min(ans,dp[(1<<N)-1][i])
print(ans if ans<INF else -1)
