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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (n); ++i )

using namespace std;
typedef long long ll;

int main(){
    int N; cin >> N;
    vector<int> in_p,in_q;
    int p,q;
    rep (i,N){
        int tmp;
        cin >> tmp;
        in_p.push_back(tmp);
        p += pow(10,N-1-i) * in_p[i];
    }
    //cout << p << endl;
    rep (i,N){
        int tmp;
        cin >> tmp;
        in_q.push_back(tmp);
        q += pow(10,N-1-i) * in_q[i];
    }
    //cout << q << endl;
    vector<int> v(N);
    vector<int> per_list;
    iota(v.begin(), v.end(), 1);       // v に 1, 2, ... N を設定
    do {
        int tmp = 0;
            rep (i,N){
                if (i > 0){
                    tmp += pow(10,i) * v[i];
                }
                else {
                    tmp += v[i];
                }
            }
            per_list.push_back(tmp);
    } while( next_permutation(v.begin(), v.end()) );
    sort(per_list.begin(),per_list.end());
    int p_count = 0;
    int q_count = 0;

    for (const auto v:per_list){
        if (p > v){
            p_count++;
        }
        if (q > v){
            q_count++;
        }
    }
    int tmp = abs(q_count-p_count);
    cout << tmp << endl;
    return 0;
}