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 <stdlib.h>

int factorial(int a[],int n);

int factorial(int a[],int n){
    int ans=0,nn=1;
    
    for(int i=2;i<=n-1;i++){nn*=i;}
    
    for(int i=0;i<n-1;i++){
        int count=0;
        for(int j=i+1;j<n;j++){
            if(a[i]<a[j]){count++;}
        }
        ans+=count*nn;
        nn=nn/(n-1-i);
    }
    return ans;
}

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