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


def main():
    import sys
    input = sys.stdin.readline
    n = int(input())
    a = tuple(map(int,input().split()))
    b = tuple(map(int,input().split()))

    dg = 20

    pp = 20

    dp = [10**5]*(10**7)

    new = set([])
    for i in range(n):
        s = (1<<i)
        dp[s*pp+i] = i
        new.add(s*dg+i)

    for k in range(n-1):
        tank = set([])
        if len(new) == 0:
            break
        for popelt in new:
            s,idx = popelt//dg,popelt%dg
            if (k-idx)%2 == 0:
                fr = a[idx]
            else:
                fr = b[idx]
            cnt = 0
            for j in range(n):
                if (s>>j)&1 == 0:
                    if (j-k)%2 == 1:
                        val = a[j]
                    else:
                        val = b[j]
                    if val >= fr and dp[(s+(1<<j))*pp+j] > dp[s*pp+idx]+j-cnt:
                        dp[(s+(1<<j))*pp+j] = dp[s*pp+idx]+j-cnt
                        tank.add((s+(1<<j))*dg + j)
                else:
                    cnt += 1
        new = tank
    
    res = 10**5
    s = (2**n-1)*pp
    for i in range(n):
        if res > dp[s+i]:
            res = dp[s+i]
    if res == 10**5:
        print(-1)
    else:
        print(res)


if __name__ == '__main__':
    main()