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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

int rad(int a);
int main(void)
{
    int n;
    scanf("%d", &n);
    int p[n], q[n];
    int i, ans;
    for (i = 0; i < n; i++) {
        scanf("%d", &p[i]);
    }
    for (i = 0; i < n; i++) {
        scanf("%d", &q[i]);
    }
    int pn = 0, qn = 0;
    for (i = 0; i < n; i++) {
        pn += (p[i] - 1) * rad(n - i  - 1);
        qn += (q[i] - 1) * rad(n - i - 1);

        for (int j = i + 1; j < n; j++) {
            if (p[j] > p[i]) {
                --p[j];
            }
            if (q[j] > q[i]) {
                --q[j];
            }
        }
    }


    printf("%d\n", abs(pn - qn));
    return 0;
}

int rad(int a)
{
    if (a == 1 || a == 0) {
        return 1;
    } else {
        return a * rad(a - 1);
    }
}