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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;

fn main() {
    input! {
        n: i32,
        p: [i32; n],
        q: [i32; n]
    }

    let n_us = n as usize;
    let mut sum1: i32 = 0;
    let mut sum2: i32 = 0;
    let mut vec1: Vec<i32> = Vec::new();
    let mut vec2: Vec<i32> = Vec::new();

    for i_us in 0..n_us {
        let i = i_us as i32;
        let c1 = less_used_count(p[i_us], &vec1);
        let c2 = less_used_count(q[i_us], &vec2);
        sum1 += (p[i_us] - 1 - c1) * factorial(n - i - 1) + p[i_us];
        sum2 += (q[i_us] - 1 - c2) * factorial(n - i - 1) + q[i_us];

        vec1.push(p[i_us]);
        vec2.push(q[i_us]);
    }

    let ans = sum1 - sum2;

    println!("{}", ans.abs());
}

fn factorial(n: i32) -> i32 {
    if n == 0 || n == 1 {
        return 1;
    };
    return n * factorial(n - 1);
}

fn less_used_count(target: i32, vec: &[i32]) -> i32 {
    let mut count = 0;

    for &v in vec {
        if v < target {
            count += 1;
        }
    }

    return count;
}