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 (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

int main(void){
    int n;
    scanf("%d", &n);
    int i, j;
    int p[8], q[8];
    for(i = 0; i < n; i++){scanf("%d", &p[i]);}
    for(i = 0; i < n; i++){scanf("%d", &q[i]);}
    for(i = 0; i < n; i++){p[i]--; q[i]--;}
    
    int a_temp[8];

    for(i = 0; i < n; i++){
        a_temp[i] = i;
    }
    int cnt = 1;
    int a, b;
    int flag_a = 0, flag_b = 0;
    int flag;
    int go_next;
    while(1){
        for(i = 0; i < n; i++){
            if(a_temp[i] != p[i]){
                break;
            }
            if(i == n - 1){
                a = cnt;
                flag_a = 1;
            }
        }

        for(i = 0; i < n; i++){
            if(a_temp[i] != q[i]){
                break;
            }
            if(i == n - 1){
                b = cnt;
                flag_b = 1;
            }
        }
        
        if(flag_a == 1 && flag_b == 1){
            break;
        }

        go_next = 0;
        //a_tempを1つ進める処理を書く
        do{
            a_temp[n - 1] += 1;
            for(i = 1; i < n; i++){
                if(a_temp[n - i] == n){
                    a_temp[n - i] = 0;
                    a_temp[n - i - 1]++;
                }
            }
            for(i = 0; i < n; i++){
                flag = 0;
                for(j = 0; j < n; j++){
                    if(a_temp[j] == i){
                        flag = 1;
                        break;
                    }
                }
                if(flag == 0){
                    break;
                }
                if(i == n - 1){
                    go_next = 1;
                }
            }
        }while(go_next == 0);
        cnt++;
    }
    printf("%d\n", a > b ? a - b : b - a);
    
    return 0;

}