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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
int kaijou(int);
int main(void){
  int N,i,j,p[1000],s[1000],a=0,b=0,pc=0,sc=0;
  scanf("%d",&N);
  for(i=0;i<N;i++){
    scanf("%d",&p[i]);
  }
   for(i=0;i<N;i++){
    scanf("%d",&s[i]);
  }
  
  
  for(i=0;i<N-2;i++){
    for(j=i+1;j<N;j++){
      if(p[i]>p[j])pc++;
    }
    if(pc>0){
  a+=pc*kaijou(N-i-1);
    }
    pc=0;
  }

  if(p[N-2]<p[N-1])a++;
  else a+=2;
  
   
  for(i=0;i<N-2;i++){
    for(j=i+1;j<N;j++){
      if(s[i]>s[j])sc++;
    }
    if(sc>0){
  b+=sc*kaijou(N-i-1);}
    sc=0;
  }
  if(s[N-2]<s[N-1])b++;
  else b+=2;
  
  printf("%d %d\n",a,b);
  
  printf("%d",abs(a-b));
  return 0;
}

int kaijou(int M){
if(M>1)return M*kaijou(M-1);
  else return 1;
}