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>
int kaijyou(int x){
  int sum=1;
  for(int i=1;i<=x;i++){
    sum=sum*(i);
  }
  return sum;
}
int main(){
  int n;
  scanf("%d",&n);
  int p[n],q[n],count[n-1],a=1,b=1,kai;
  /*一回目*/
  for(int i=0;i<n;i++){
    scanf("%d",&p[i]);
    count[i]=1;
  }
  for(int i=0;i<n-1;i++){
    for(int j=i+1;j<n;j++){
      if(p[i]>p[j])
        count[i]++;
    }
  }
  for(int i=0;i<n-1;i++){
    a=a+kaijyou(i+1)*(count[n-2-i]-1);
  }
  
  /*二回目*/
  for(int i=0;i<n;i++){
    scanf("%d",&q[i]);
    count[i]=1;
  }
  for(int i=0;i<n-1;i++){
    for(int j=i+1;j<n;j++){
      if(q[i]>q[j])
        count[i]++;
    }
  }
  for(int i=0;i<n-1;i++){
    b=b+kaijyou(i+1)*(count[n-2-i]-1);
  }
  
  printf("%d",abs(a-b));
  return 0;
}