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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
#include <stack>
#include <functional>

#ifdef LOCAL
    #define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
    #define eprintf(...) 42
#endif

#define rep(i, n) for(int i = 0, i##_len = (n); i < i##_len; ++i)
#define repp(i, m, n) for(int i = m, i##_len = (n); i < i##_len; ++i)
#define reprev(i, n) for(int i = (n-1LL); i >= 0; --i)
#define all(x) (x).begin(), (x).end()
template <class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template <class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return 1; } return 0; }
template <class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair <int,int> P;
typedef long double ld;

int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<int> p(n), q(n), r(n);
    rep (i, n) r[i] = i + 1;
    rep (i, n) cin >> p[i];
    rep (i, n) cin >> q[i];
    int a = 0, b = 0;
    while (true) {
        bool c = true;
        rep (i, n) if (p[i] != r[i]) c = false;
        if (c) break;
        a++;
        next_permutation(all(r));
    }
    rep (i, n) r[i] = i + 1;
    while (true) {
        bool c = true;
        rep (i, n) if (q[i] != r[i]) c = false;
        if (c) break;
        b++;
        next_permutation(all(r));
    }
    cout << abs(a - b) << "\n";
    return 0;
}