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 3.8.0) (judge verdict: Accepted)

#include <inttypes.h>
#include <math.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

intmax_t N;
char P[9] = {}, Q[9] = {};
intmax_t ans = 0;
char tmp[9] = {};
char p[8 * 7 * 6 * 5 * 4 * 3 * 2 * 1][9] = {};
size_t idx = 0;
bool used[8] = {};

void f(intmax_t pos)
{
    if (pos >= N) {
        strcpy(&p[idx][0], tmp);
        idx++;
        return;
    }
    for (intmax_t i = 1; i <= N; i++) {
        if (used[i - 1])
            continue;
        tmp[pos] = i + '0';
        used[i - 1] = true;
        f(pos + 1);
        used[i - 1] = false;
    }
}

int main(void)
{
    scanf("%jd", &N);
    for (size_t i = 0; i < N; i++) {
        intmax_t pi;
        scanf("%jd", &pi);
        P[i] = pi + '0';
    }
    for (size_t i = 0; i < N; i++) {
        intmax_t qi;
        scanf("%jd", &qi);
        Q[i] = qi + '0';
    }

    f(0);

    size_t a, b;
    for (size_t i = 0; p[i][0]; i++) {
        if (strcmp(P, &p[i][0]) == 0) {
            a = i;
        }
        if (strcmp(Q, &p[i][0]) == 0) {
            b = i;
        }
    }

    ans = a - b;
    if (ans < 0)
        ans = -ans;

    printf("%jd\n", ans);

    return 0;
}