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>

long getFactorial(int number)
{

    long ans = 1;
    if (number == 0)
    {
        ans = 0;
    }
    else
    {
        while (number > 0)
        {
            ans = ans * number;
            number--;
        }
    }

    return ans;
}

int main(void)
{

    int i_inputPermutationSize;
    int ia_inputFirstPermutation[8];
    int ia_inputSecondPermutation[8];
    scanf("%d", &i_inputPermutationSize);
    for (int index = 0; index < i_inputPermutationSize; index++)
    {
        scanf("%d", &ia_inputFirstPermutation[index]);
    }
    for (int index = 0; index < i_inputPermutationSize; index++)
    {
        scanf("%d", &ia_inputSecondPermutation[index]);
    }

    long l_firstPermutationNumber = 1;
    long l_secondPermutationNumber = 1;
    for (int index = 0; index < i_inputPermutationSize; index++)
    {
        int counter = ia_inputFirstPermutation[index];
        for (int index2 = index - 1; index2 >= 0; index2--)
        {
            if (ia_inputFirstPermutation[index] > ia_inputFirstPermutation[index2])
            {
                counter--;
            }
        }
        for (int count = 1; count < counter; count++)
        {
            l_firstPermutationNumber += getFactorial(i_inputPermutationSize - (index + 1));
        }
    }
    for (int index = 0; index < i_inputPermutationSize; index++)
    {
        int counter = ia_inputSecondPermutation[index];
        for (int index2 = index - 1; index2 >= 0; index2--)
        {
            if (ia_inputSecondPermutation[index] > ia_inputSecondPermutation[index2])
            {
                counter--;
            }
        }
        for (int count = 1; count < counter; count++)
        {
            l_secondPermutationNumber += getFactorial(i_inputPermutationSize - (index + 1));
        }
    }

    long answer = 0;
    if (l_firstPermutationNumber > l_secondPermutationNumber)
    {
        answer = l_firstPermutationNumber - l_secondPermutationNumber;
    }
    else if (l_firstPermutationNumber < l_secondPermutationNumber)
    {
        answer = l_secondPermutationNumber - l_firstPermutationNumber;
    }

    printf("%ld", answer);

    return 0;
}