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: Accepted)

#include "bits/stdc++.h"

// Begin Header {{{
using namespace std;

#ifndef DEBUG
#define dump(...)
#endif

#define all(x) x.begin(), x.end()
#define rep(i, n) for (intmax_t i = 0, i##_limit = (n); i < i##_limit; ++i)
#define reps(i, b, e) for (intmax_t i = (b), i##_limit = (e); i <= i##_limit; ++i)
#define repr(i, b, e) for (intmax_t i = (b), i##_limit = (e); i >= i##_limit; --i)
#define var(Type, ...) Type __VA_ARGS__; input(__VA_ARGS__)

constexpr size_t    operator""_zu(unsigned long long value) { return value; };
constexpr intmax_t  operator""_jd(unsigned long long value) { return value; };
constexpr uintmax_t operator""_ju(unsigned long long value) { return value; };

constexpr int INF = 0x3f3f3f3f;
constexpr intmax_t LINF = 0x3f3f3f3f3f3f3f3f_jd;

template <class T>
using MaxHeap = priority_queue<T, vector<T>, less<T>>;
template <class T>
using MinHeap = priority_queue<T, vector<T>, greater<T>>;

inline void input() {}
template <class Head, class... Tail>
inline void input(Head&& head, Tail&&... tail) {
    cin >> head;
    input(forward<Tail>(tail)...);
}

template <class T>
inline void input(vector<T> &vec) {
    for (auto &e: vec) {
        cin >> e;
    }
}

template <class T>
inline void input(vector<vector<T>> &mat) {
    for (auto &vec: mat) {
        input(vec);
    }
}

inline void print() { cout << "\n"; }
template <class Head, class... Tail>
inline void print(Head&& head, Tail&&... tail) {
    cout << head;
    if (sizeof...(tail)) {
        cout << ' ';
    }
    print(forward<Tail>(tail)...);
}

template <class T>
inline ostream& operator<<(ostream &strm, const vector<T> &vec) {
    static constexpr const char *delim[] = {" ", ""};
    for (const auto &e: vec) {
        strm << e << delim[&e == &vec.back()];
    }
    return strm;
}

template <class T>
inline vector<T> make_v(const T &initValue, size_t sz) {
    return vector<T>(sz, initValue);
}

template <class T, class... Args>
inline auto make_v(const T &initValue, size_t sz, Args... args) {
    return vector<decltype(make_v<T>(initValue, args...))>(sz, make_v<T>(initValue, args...));
}

template <class Func>
class FixPoint : Func {
public:
    explicit constexpr FixPoint(Func&& f) noexcept : Func(forward<Func>(f)) {}

    template <class... Args>
    constexpr decltype(auto) operator()(Args&&... args) const {
        return Func::operator()(*this, std::forward<Args>(args)...);
    }
};

template <class Func>
static inline constexpr decltype(auto) makeFixPoint(Func&& f) noexcept {
    return FixPoint<Func>{forward<Func>(f)};
}

template <class T>
inline bool chmax(T &a, const T &b) noexcept {
    return b > a && (a = b, true);
}

template <class T>
inline bool chmin(T &a, const T &b) noexcept {
    return b < a && (a = b, true);
}

template <class T>
inline T diff(const T &a, const T &b) noexcept {
    return a < b ? b - a : a - b;
}
// }}}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    var(size_t, N);

    vector<size_t> p(N);
    input(p);

    vector<size_t> q(N);
    input(q);

    vector<size_t> v(N);
    iota(all(v), 1);

    vector<vector<size_t>> buff;
    do {
        buff.emplace_back(v);
    } while (next_permutation(all(v)));

    sort(all(buff));

    int a = 0, b = 0;
    rep(i, buff.size()) {
        if (buff[i] == p) {
            a = i;
        }
        if (buff[i] == q) {
            b = i;
        }
    }

    print(diff(a, b));

    return 0;
}