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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

typedef long long ll;

int max(int a, int b){
  if(a > b) return a;
  else return b;
}
int min(int a, int b){
  if(a < b) return a;
  else return b;
}

int sort(const void *a ,const void *b){
  return *(int*)a - *(int*)b;
}

  int ans[10];
  int p[9];
  int q[9];
  int n;
  int flag[10] = {0};
  int a,b;
  int count = 0;

void dfs(int digit){

  if(digit == n){
    count ++;
    int flaga = 1,flagb = 1;
    for(int i = 0; i < n; ++i){
      if(ans[i] != p[i]) flaga = 0;
      if(ans[i] != q[i]) flagb = 0;
    }
    if(flaga == 1) a = count;
    if(flagb == 1) b = count;
    return;
  }

  for(int i = 1; i <= n; ++i){
    ans[digit] = i;
    if(flag[i] == 0){
      flag[i] = 1;
      dfs(digit+1);
      flag[i] = 0;
    }
  }

}

int main(){

  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);
  }

  dfs(0);

  printf("%d",abs(a-b));

  return 0;
}