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<stdlib.h>
#include<stdio.h>
#include<string.h>
void swap(int *x,int *y){int tmp=*x;*x=*y;*y=tmp;}
int asc_cmp(const void* X,const void* Y){int x=*(int*)X,y=*(int*)Y;if(x<y)return -1;else if(x==y)return 0;else return 1;}
//next_permutation
//from https://qiita.com/Nikkely/items/0ddca51b3c0e60afbaab
int next_permutation(int *a,int n){
    int x=-1,y=-1;
    for(int i=n-2;i>=0;i--)if(a[i]<a[i+1]){x=i;break;}
    for(int i=n-1;x!=-1&&i>=0;i--)if(a[x]<a[i]){y=i;break;}
    if(x==-1||y==-1)return 1;
    swap(&a[x],&a[y]);
    qsort(&a[x+1],n-(x+1),sizeof(int),asc_cmp);
    return 0;
}

int main(){
    int n;scanf("%d",&n);
    int p[20];for(int i=0;i<n;i++)scanf("%d",&p[i]);
    int q[20];for(int i=0;i<n;i++)scanf("%d",&q[i]);

    char p_str[20],q_str[20];
    for(int i=0;i<n;i++)p_str[i]=p[i]+'0';
    for(int i=0;i<n;i++)q_str[i]=q[i]+'0';
    p_str[n]=q_str[n]='\0';

    int a[n];for(int i=1;i<=n;i++)a[i-1]=i;

    int index[2];
    static char list[100000][20];
    for(int i=0;1;i++){
        for(int j=0;j<n;j++)list[i][j]='0'+a[j];
        list[i][n]='\0';
        if(!strcmp(list[i],p_str))index[0]=i;
        if(!strcmp(list[i],q_str))index[1]=i;
        if(next_permutation(a,n))break;
    }
    
    printf("%d\n",abs(index[0]-index[1]));
    return 0;
}