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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main() {
    int N;
    cin >> N;
    vector<int> num(N);
    vector<int> input_P(N);
    vector<int> input_Q(N);
    int order_P = 0;
    int order_Q = 0;
    for(int i = 0; i < N; i++) {
        num[i] = i + 1;
    }
    for(int i = 0; i < N; i++) {
        cin >> input_P[i];
    }
    for(int i = 0; i < N; i++) {
        cin >> input_Q[i];
    }
    int count = 0;
    do {
        count++;
        for(int i = 0; i < N; i++) {
            if(num[i]  != input_P[i]) {
                break;
            }
            if(i == N - 1) {
                order_P = count;
            }
        }
        for(int i = 0; i < N; i++) {
            if(num[i]  != input_Q[i]) {
                break;
            }
            if(i == N - 1) {
                order_Q = count;
            }
        }
    } while(next_permutation(num.begin(), num.end()));
    cout << abs(order_P - order_Q) << endl;
    return 0;
}