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;
#define  rep(i, n) for(int i = 0; i < n; i++) 
#define  all(x) (x).begin(),(x).end()     // 昇順ソート
#define  rall(v) (v).rbegin(), (v).rend() // 降順ソート
#define  INF 1LL << 60
typedef long long int LL;
typedef long long int ll;
#define pll pair<ll, ll>
#define F first
#define S second
const int MOD = 1000000007;
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return true; } return false; }
//sort(all(x))とするとソートできるよ
// 10^x は pow(10,(x)) 
// 任意のlogは　対数の底の変換を使う    log(N) / log(任意の底)

int main(){
    int N;cin >> N;
    vector<int> P(N);
    vector<int> Q(N);
    vector<int> ar(N);
    rep(i,N)cin >> P[i];
    rep(i,N)cin >> Q[i];
    rep(i,N)ar[i] = i+1;

    int pn = 0;
    int qn = 0;
    int ct1 = 0;
    int ct2 = 0;
    do{
        int tmp = 0;
        int p = 0;
        for(int i=N-1; i>=0; i--){
            tmp += ar[i] * pow(10,i);
            p += P[i] * pow(10,i);
        }
 //       cout << tmp << " " << p << endl;
        if(tmp == p)pn = ct1;
        ct1++;
    }while(next_permutation(all(ar)));

    do{
        int tmp = 0;
        int q = 0;
        for(int i=N-1; i>=0; i--){
            tmp += ar[i] * pow(10,i);
            q += Q[i] * pow(10,i);
        }
        if(tmp == q)qn = ct2;
        ct2++;
    }while(next_permutation(all(ar)));

 //   cout << pn << " " << qn << endl;
    cout << abs(pn-qn) << endl;
}