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<vector>
#include<unordered_map>
#include<algorithm>
using namespace std;
string encode(vector<int> v){
    string res = "";
    for(int i = 0; i<v.size(); i++){
        res += to_string(v[i]) + "#";
    }
    return res;
}
void permute(int curr, vector<int>& v, vector<vector<int> >& res){
    if(curr == v.size()){
        res.push_back(v);
        return;
    }
    for(int i= curr; i<v.size(); i++ ){
        swap(v[i], v[curr]);
        permute(curr + 1, v, res);
        swap(v[i], v[curr]);
    }
}
int main(){
    int n;
    cin>>n;
    vector<int> v1(n);
    for(int i = 0; i<n; i++){
        cin>>v1[i];
    }
    string s1 = encode(v1);
    vector<int> v2(n);
    for(int i = 0; i<n; i++){
        cin>>v2[i];
    }
    string s2 = encode(v2);

    vector<int> v;
    for(int i= 1; i<=n; i++){
       v.push_back(i); 
    }
    unordered_map<string , int> m;
    vector<vector<int> > res;
    int num = 1;
    permute(0, v, res);
    sort(res.begin(), res.end());
    for(int i = 0; i<res.size(); i++){
        string x = encode(res[i]);
        m[x] = i+1;
    }
    cout<<abs(m[s1]- m[s2])<<endl;
    return 0;
}