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 text_io::*;
// use std::collections::*;
// use std::process::exit;

fn main() {
    let n:usize = read!();
    let mut p:Vec<usize> = Vec::new();
    for _ in 0..n {
        let kk:usize = read!();
        p.push(kk);
    }
    let mut q:Vec<usize> = Vec::new();
    for _ in 0..n {
        let kk:usize = read!();
        q.push(kk);
    }
    let mut pp:usize = 0;
    for i in 0..n-1 {
        let mut ppp:usize = 1;
        for j in 1..n-i {
            ppp *= j;
        }
        pp += (p[i]-1)*ppp;
        for j in i+1..n-1 {
            if p[i]<p[j] {
                p[j]-=1;
            }
        }
    }
    let mut qq:usize = 0;
    for i in 0..n-1 {
        let mut qqq:usize = 1;
        for j in 1..n-i {
            qqq *= j;
        }
        qq += (q[i]-1)*qqq;
        for j in i+1..n-1 {
            if q[i]<q[j] {
                q[j]-=1;
            }
        }
    }
    println!("{}",(pp as isize-qq as isize).abs());
}