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 JavaScript (node.js v5.12) (judge verdict: Accepted)

function Main(input) {
    input = input.trim().split("\n").map(function(x) { return x.split(" ")});
    var N = parseInt(input[0], 10);
    var vec_P = input[1].map(e => parseInt(e, 10));
    var vec_Q = input[2].map(e => parseInt(e, 10));
    var pi, qi, a, b;
    a = b = 1;
    for (var i = 0; i < (N - 1); i++){
    	pi = vec_P[0];
    	qi = vec_Q[0];
    	
    	vec_P.shift();
    	vec_Q.shift();
    	a += (N - (i + 1) - countAbove(pi, vec_P)) * kaijo(N - (i + 1));
    	b += (N - (i + 1) - countAbove(qi, vec_Q)) * kaijo(N - (i + 1));
    }
	console.log(Math.abs(a-b));
}

function countAbove(num, vec){
	var ans = 0;
	for (var i = 0; i < vec.length; i++){
		if (vec[i] > num) ans++;
	} 
	return ans;	
}

function kaijo(num){
	var ans = 1;
	for (var i = 2; i <= num; i++){
		ans *= i;
	} 
	return ans;	
}

Main(require("fs").readFileSync("/dev/stdin", "utf8"));