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>
using namespace std;
using ll = long long;
using ld = long double;
using VI = vector<int>;
using VL = vector<ll>;
template<class T> using PQ = priority_queue<T, vector<T>, greater<T>>;
#define FOR(i,a,n) for(int (i)=(a);(i)<(n);++(i))
#define eFOR(i,a,n) for(int (i)=(a);(i)<=(n);++(i))
#define rFOR(i,a,n) for(int (i)=(n)-1;(i)>=(a);--(i))
#define erFOR(i,a,n) for(int (i)=(n);(i)>=(a);--(i))
#define each(i, a) for(auto &i : a)
#define SORT(i) sort((i).begin(),(i).end())
#define rSORT(i,a) sort((i).begin(),(i).end(),(a))
#define all(i) (i).begin(),(i).end()
#define out(y,x) ((y) < 0 || h <= (y) || (x) < 0 || w <= (x))
#define line cout << "------------------------\n" 
#define ENDL(i,n) ((i) == (n) - 1 ? "\n" : " ")
#define stop system("pause") //comment out this on AOJ.
constexpr ll INF = 1000000000;
constexpr ll LLINF = 1LL << 60;
constexpr ll mod = 1000000007;
constexpr ll MOD = 998244353;
constexpr ld eps = 1e-10; //1e-9?
constexpr ld pi = 3.1415926535897932;
template<class T>inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; }return false; }
template<class T>inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; }return false; }
inline void init() { cin.tie(nullptr); cout.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); }
template<class T>inline istream& operator>>(istream& is, vector<T>& v) { for (auto& elemnt : v)is >> elemnt; return is; }
template<class T, class U>inline istream& operator>>(istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }
template<class T>inline vector<T> vec(size_t a) { return vector<T>(a); }
template<class T>inline vector<T> defvec(T def, size_t a) { return vector<T>(a, def); }
template<class T, class... Ts>inline auto vec(size_t a, Ts... ts) { return vector<decltype(vec<T>(ts...))>(a, vec<T>(ts...)); }
template<class T, class... Ts>inline auto defvec(T def, size_t a, Ts... ts) { return vector<decltype(defvec<T>(def, ts...))>(a, defvec<T>(def, ts...)); }

int main() {
    init();

    int n; cin >> n;
    VI p(n), q(n); cin >> p >> q;
    VI a(n);
    FOR(i, 0, n)a[i] = i + 1;

    int cnt = 1, x = -1, y = -1;

    do {
        bool P = true, Q = true;
        FOR(i, 0, n) {
            if (a[i] != p[i])P = false;
            if (a[i] != q[i])Q = false;
        }
        if (P)x = cnt;
        if (Q)y = cnt;
        ++cnt;
    } while (next_permutation(all(a)));

    cout << abs(x - y) << "\n";
}