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
#!/usr/bin/env python3
import sys
input = sys.stdin.readline

n = int(input())
a = [int(item) for item in input().split()]
b = [int(item) for item in input().split()]

odd_bit = 0b0101010101010101010101
eve_bit = 0b1010101010101010101010
full_bit = 2**n - 1

odd_num = (n + 1) // 2
ans = 10**9 

for i in range(2**n):
    invi = full_bit ^ i
    odd = bin(i & odd_bit).count("1") + bin(invi & eve_bit).count("1")
    if odd != odd_num:
        continue
    odd_card = []
    eve_card = []
    for j in range(n):
        if i & 1 << j:
            if j % 2 == 0:
                odd_card.append((a[j], j))
            else:
                eve_card.append((a[j], j))
        else:
            if j % 2 == 0:
                eve_card.append((b[j], j))
            else:
                odd_card.append((b[j], j))
    odd_card.sort()
    eve_card.sort()
    odd_itr = 0
    eve_itr = 0
    curr = 0
    val = 0
    ok = True
    perm = []
    for i in range(n):
        if i % 2 == 0:
            num, index = odd_card[odd_itr]
            perm.append(index)
            odd_itr += 1
        else:
            num, index = eve_card[eve_itr]
            perm.append(index)
            eve_itr += 1
        if num < curr:
            ok = False
            break
        curr = num
    if ok:
        val = 0
        visited = [0] * n
        itr = 0
        for item in perm:
            visited[item] = 1
            val += itr - sum(visited[:item])
            itr += 1
        ans = min(ans, val)

if ans == 10**9:
    print(-1)
else:
    print(ans)