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 <string>
#include <vector>
#include <algorithm>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <iomanip>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstring>
#include <climits>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define SORT(c) sort((c).begin(), (c).end())

typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> V;
typedef map<int, int> M;

constexpr ll INF = 1e18;
constexpr ll MOD = 1e9 + 7;
constexpr double PI = 3.14159265358979323846;
constexpr int di[] = {0, 0, 1, -1};
constexpr int dj[] = {1, -1, 0, 0};

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n;
    V p, q;

    cin >> n;
    REP(i, n)
    {
        int b;
        cin >> b;
        p.push_back(b);
    }
    REP(i, n)
    {
        int b;
        cin >> b;
        q.push_back(b);
    }

    if (p == q)
    {
        cout << 0 << endl;
        return 0;
    }

    if (p > q)
        swap(p, q);

    V v(n);
    REP(i, n)
    v[i] = i + 1;

    int cnt = 1;

    do
    {
        if (p < v && v < q)
            cnt++;
    } while (next_permutation(v.begin(), v.end()));

    cout << cnt << endl;

    return 0;
}