from collections import defaultdict, deque, Counter

# ２つの数列　隣接swap　一致させる　要素に重複なし


# 転倒数
def mergeCount(A):
    cnt = 0
    n = len(A)
    if n > 1:
        A1 = A[:n >> 1]
        A2 = A[n >> 1:]
        cnt += mergeCount(A1)
        cnt += mergeCount(A2)
        i1 = 0
        i2 = 0
        for i in range(n):
            if i2 == len(A2):
                A[i] = A1[i1]
                i1 += 1
            elif i1 == len(A1):
                A[i] = A2[i2]
                i2 += 1
            elif A1[i1] <= A2[i2]:
                A[i] = A1[i1]
                i1 += 1
            else:
                A[i] = A2[i2]
                i2 += 1
                cnt += n//2 - i1
    return cnt


n = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))

if Counter(A) != Counter(B):
    print(-1)
    exit()

# print(A)
# print(B)

# A[i] = B[s_i] となるような配列 S (s_0, s_1, ,,,)を考える
# A[i], B[i]で要素の値が等しいものが複数あるときは、i < j, s_i < s_jとなるように決める
# 決定したSに対して転倒数を求める

# Bの要素の値ごとの位置を調べる
pos = defaultdict(deque)
for itr, b in enumerate(B):
    pos[b].append(itr)

S = []
for a in A:
    p = pos[a].popleft()
    S.append(p)
# print(S)

print(mergeCount(S))