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
# coding: utf-8
# Your code here!
import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline
read = sys.stdin.read

#n = int(input)
#h,w = [int(i) for i in readline().split()]
#n,k,s = [int(i) for i in read().split()]

n = int(input())
a = [int(i)-1 for i in readline().split()]
b = [int(i)-1 for i in readline().split()]

N = 1<<n
ans = INF = 10**9
for mask in range(N):
    bucket = [[] for _ in range(50)]
    for i in range(n):
        bucket[a[i] if mask&(1<<i) else b[i]].append(i)
    
    cnt = 0
    res = []
    #print(bucket)
    #if mask == 3: print(bucket[:5])
    for bi in bucket:
        if not bi: continue
        l = len(bi)
        o = [i for i in bi if ((mask>>i)^i)&1==0]
        e = [i for i in bi if ((mask>>i)^i)&1==1]
        odd = len(o)
        even = len(e)
        #if mask == 3: print(cnt,bi,o,e)
        if l%2 == 0:
            if odd != even: break
        else:
            if cnt%2==0:
                if even != odd+1: break
            else:
                if odd != even+1: break
        
        #偶奇を突破したら
        for i in range(l):
            res.append(e[i//2] if (i+cnt)%2==0 else o[i//2])
        cnt += l
        
        
    else: #うまくいったら
        num = 0
        for i in range(n):
            for j in range(i+1,n):
                if res[i] > res[j]: num += 1
        #print(mask,num,res)
        ans = min(ans,num)                

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


