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
INF = 10000
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))

def popcount(x):
    x = x - ((x >> 1) & 0x5555555555555555)
    x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)
    x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f
    x = x + (x >> 8)
    x = x + (x >> 16)
    x = x + (x >> 32)
    return x & 0x0000007f

dp = [[INF for i in range(N)] for j in range(1 << N)]

for i in range(N):
    dp[1 << i][i] = i

for i in range(1 << N):
    for j in range(N):
        if i >> j & 1:
            if (popcount(i) - j) % 2 != 0:
                num = A[j]
            else:
                num = B[j]
            cnt = 0
            for k in range(N):
                if not (i >> k & 1):
                    if (popcount(i) - k) % 2 == 0:
                        num2 = A[k]
                    else:
                        num2 = B[k]
                    if num <= num2:
                        dp[i | 1 << k][k] = min(dp[i | 1 << k][k], dp[i][j] + cnt)
                    cnt += 1

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

if ans == INF:
    print(-1)
else:
    print(ans)