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())
As = list(map(int, input().split()))
Bs = list(map(int, input().split()))

INF = 1<<20

dp = [[INF] * 55 for i in range(1<<n)]
dp[0][0] = 0

for b in range(1<<n):
    # 使ったカードの枚数
    bitcnt = bin(b).count("1")
    cur = bitcnt 
    rest = [] 
    for i in range(n):
        # i番目のbitが立っているかの判定
        if not (b & (1 << i)):
            rest.append((i,cur))
            cur += 1
    
    for s in range(51):
        if dp[b][s] >= INF:
            continue
        # 残ったカードで更新
        for i,cur in rest:
            if abs(bitcnt - i) % 2 == 0:
                num = As[i]
            else:
                num = Bs[i]
            if num >= s:
                d = abs(bitcnt - cur) 
                dp[b|(1<<i)][num] = min(dp[b|(1<<i)][num], dp[b][s] + d)
            

ans = min(dp[(1<<n)-1])
print(ans if ans != INF else -1)    
        
    
