Problem:
Score : 300 points

Problem Statement

We have two permutations P and Q of size N (that is, P and Q are both rearrangements of (1,~2,~...,~N)).

There are N! possible permutations of size N. Among them, let P and Q be the a-th and b-th lexicographically smallest permutations, respectively. Find |a - b|.

Notes

For two sequences X and Y, X is said to be lexicographically smaller than Y if and only if there exists an integer k such that X_i = Y_i~(1 \leq i < k) and X_k < Y_k.

Constraints

2 \leq N \leq 8

P and Q are permutations of size N.

Input

Input is given from Standard Input in the following format:

N
P_1 P_2 ... P_N
Q_1 Q_2 ... Q_N

Output

Print |a - b|.

Sample Input 1

3
1 3 2
3 1 2

Sample Output 1

3

There are 6 permutations of size 3: (1,~2,~3), (1,~3,~2), (2,~1,~3), (2,~3,~1), (3,~1,~2), and (3,~2,~1). Among them, (1,~3,~2) and (3,~1,~2) come 2-nd and 5-th in lexicographical order, so the answer is |2 - 5| = 3.

Sample Input 2

8
7 3 5 4 2 1 6 8
3 8 2 5 4 6 7 1

Sample Output 2

17517

Sample Input 3

3
1 2 3
1 2 3

Sample Output 3

0

Solution in PyPy3 (2.4.0) (judge verdict: Runtime Error)

N = int(input())
P = list(map(int,input().split()))
Q = list(map(int,input().split()))

def kaijou(n):
    x = 1
    for i in range (1,n+1):
        x *= i
    return x

kaijou_ls = [0]*N

for i in range(N):
    kaijou_ls[i] = kaijou(i+1)

#dp
def count_order(A,k):
    if k==1:
        return 1
    else:
        B = A[N-k:]
        b = B[0]
        C = B.copy()
        C.sort()
        order = C.index(b)*kaijou_ls[k-2]

        return order+count_order(A,k-1)

print(abs(count_order(P,N)-count_order(Q,N)))