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 (Clang 10.0.0) (judge verdict: Accepted)

#include <stdio.h>

int main (){
	int n,i,sw,j;
	int count;
	int jisho[8];
	int p[8];
	int q[8];
	int pc=0;
	int qc=0;
	scanf("%d",&n);
	count = 1;
	for(i=0;i<n;i++){
		jisho[i] = i+1;
	}
	for(i=0;i<n;i++){
		count = 1;
		scanf("%d",&p[i]);
		for(j=n-1-i;j>0;j--){
			count = count*j;
		}
		j=0;sw=0;
		while(sw>=0){
			if(p[i]==jisho[j]){
				pc += count*sw;
				jisho[j]= -1;
				sw = -1;
			} else if(jisho[j]>0){
					sw ++;
			}
			j++;
		}
	}
	for(i=0;i<n;i++){
		jisho[i] = i+1;
	}
	for(i=0;i<n;i++){
		count = 1;
		scanf("%d",&q[i]);
		for(j=n-1-i;j>0;j--){
			count = count*j;
		}
		j=0;sw=0;
		while(sw>=0){
			if(q[i]==jisho[j]){
				qc += count*sw;
				jisho[j]= -1;
				sw = -1;
			} else if(jisho[j]>0){
					sw ++;
			}
			j++;
		}
	}

	if(pc>qc){
		printf("%d\n",pc-qc);
	}else{
		printf("%d\n",qc-pc);
	}


	return 0;
}