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<string.h>
#include<stdlib.h>
#include<limits.h>
typedef long long int lli;

int num[110];
int frag[110];
int q[8], p[8];
lli pp, qq, cnt=0;
void fx(int n, int k)
{
    int i, f, j;
    for (i = 0; i < n; ++i)
    {
        f = frag[i];
        if (f == 0)
        {
            frag[i] = 1;
            num[k] = i + 1;
            if (k == 1)
            {
                cnt++;
                int flg1, flg2;
                flg1 = flg2 = 0;
                for (j = n; j > 0; --j)
                {
                    if(num[j]!=q[n-j]){
                        flg1 = 1;
                        break;
                    }
                }
                for (j = n; j > 0;--j){
                    if (num[j] != p[n - j])
                    {
                        flg2 = 1;
                        break;
                    }
                }
                if(flg1==0){
                    qq = cnt;
                }
                if(flg2==0){
                    pp = cnt;
                }
            }
            else
            {
                fx(n, k - 1);
            }
            frag[i] = 0;
        }
    }
}

int int_comp(const void*a,const void*b){return *(int*)a-*(int*)b;}
int lli_comp(const void*a,const void*b){return *(lli*)a-*(lli*)b;}
int main(void)
{

    int n;

    pp = qq = 0;
    scanf("%d", &n);
    for (int i = 0;i < n;i++) scanf("%d", &q[i]);
    for (int i = 0; i < n; i++) scanf("%d", &p[i]);

    for (int i = 0; i < n; i++){
        num[i] = frag[i] = 0;
    }
    fx(n, n);
    printf("%lld\n", llabs(pp-qq));
    return 0;
}