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 <algorithm>
#include <numeric>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <functional>
using namespace std;
#define int long long
#define All(v) (v).begin(),(v).end()
int dy[8] = {-1,0,1,0,-1,1,1,-1};
int dx[8] = {0,1,0,-1,1,1,-1,-1};
const int mod = 1000000007;
const int inf = mod*mod;
const int d5 = 100100;
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin>>n;
    vector<int> p(n),q(n),a(n);
    for(int i=0;i<n;i++)cin>>p[i];
    for(int i=0;i<n;i++)cin>>q[i];
    for(int i=0;i<n;i++)a[i]=i+1;
    int x,y,cnt=0;
    do{
        cnt++;
        bool P=true,Q=true;
        for(int i=0;i<n;i++){
            if(a[i]!=p[i])P=false;
            if(a[i]!=q[i])Q=false;
        }
        if(P)x=cnt;
        if(Q)y=cnt;
    }while(next_permutation(All(a)));
    cout<<abs(x-y)<<endl;

}