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;

#define fst first
#define scd second
#define PB push_back
#define MP make_pair
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define omajinai ios::sync_with_stdio(false);cin.tie(0)
#define rep(i,x) for(int i=0;i<(int)(x);++i)
#define rep1(i,x) for(int i=1;i<=(int)(x);++i)
#define rrep(i,x) for(int i=(int)x-1;i>=0;--i)
#define rrep1(i,x) for(int i=(int)x;i>=1;--i)

using ll=long long;
using ld=long double;
using vi=vector<int>;
using vvi=vector<vi>;
using pii=pair<int, int>;
using vpii=vector<pii>;

template<class T,class U>ostream&operator<<(ostream&os,const pair<T,U>p){os<<"("<<p.fst<<", "<<p.scd<<")";return os;}
template<class T>ostream&operator<<(ostream&os,const vector<T>v){rep(i,v.size()){if(i)os<<", ";os<<v[i];}return os;}
template<typename T>T&max(T&a,T&b){if(a>=b)return a;return b;}
template<typename T>T&min(T&a,T&b){if(a < b)return a;return b;}
template<typename T>bool chmax(T&a,T b){if(a < b){a=b;return true;}return false;}
template<typename T>bool chmin(T&a,T b){if(a > b){a=b;return true;}return false;}

constexpr int inf = 1e9;
constexpr ll linf = 3e18;
constexpr double eps = 1e-9;

int N;

int f[11];

// 6 8 4 2 7 5 3 1
// (5~1) ... (7!)
// 6 (7~1) ... (6!)
// 6 8 

// 1 3 2
// (0) ... (2!) 0
// 1 (2~1) 1!
// 1 3 2

int calc(vector<int> a) {
    int ret = 0;
    rep(i, N) {
        int cnt = 0;
        rep(j, i) if (a[i] > a[j]) cnt++;
        int tmp = (a[i] - 1 - cnt) * f[(N - (i+1))];
        cout << tmp << endl;
        ret += tmp;
    }

    return ret;
}

signed main()
{
    f[0] = 1;
    rep1(i, 10) f[i] = i * f[i-1];

    cin >> N;

    vector<int> P(N), Q(N);

    rep(i, N) cin >> P[i];
    rep(i, N) cin >> Q[i];

    // cout << "calc(P) = " << calc(P) << endl;
    // cout << "calc(Q) = " << calc(Q) << endl;

    int ans = calc(P) - calc(Q);
    if (ans < 0) ans = -ans;

    cout << ans << endl;
    
    return 0;
}