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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")

const int maxn = 2e5 + 20, lg = 20, inf = 1e9 + 5;

int main(){
    ios::sync_with_stdio(false);
    cout.tie(0);
    cin.tie(0);

    int n;
    cin >> n;
    int p[n], q[n];

    long long fact = 1;
    for(int i = 0; i < n; i++){
        cin >> p[i];
        fact *= (i + 1);
    }
    for(int i = 0; i < n; i++){
        cin >> q[i];
    }
    long long ans = 0;
    while(memcmp(p, q, sizeof q)){
        ans++;
        next_permutation(p, p + n);
    }
    cout << min(ans, fact - ans);
}