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)

//aea ctmr
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
//v.erase(unique(v.begin(),v.end()),v.end());
#define f(i,a,b) for( ll i = a; i < b ; i++ ) 
#define af(i,a,b) for( ll i = a; i >= b ; i--)
#define rep(i,a,b,k) for(ll i = a; i < b ; i+= k )
#define arep(i,a,b,k) for( ll i = a; i >= b ; i-= k)
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define sor(a) sort( a.begin(), a.end() )
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL)
#define PI 3.14159265358979323846
 
typedef long double ld;
typedef long long ll;
const ll N = 100500;
const ll mod = 1e9 + 7;
using namespace std;
 
#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << endl;}
 
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1<<" | ";
  __f(comma+1, args...);}
#else
#define trace(...)
#endif


int n;
int fact[9];
vector <int> xd;

int tmr(set <int> s, int x){
    int u = 1;
    for(auto _x: s){
        if(_x == x) return u;
        u++;
    }
}

int posi(){
    int ans = 1;
    set <int> s; s.clear();
    for(int i = 1; i <= n; i++) s.insert(i);

    for(int i = 0; i < n; i++){
        int pos = tmr(s, xd[i]);
        s.erase(xd[i]);
        s.erase(xd[i]);
        ans += (pos - 1) * fact[n - i - 1];

    }

    return ans;
}

int main(){
    fastio;
    int resp = 0;
    cin >> n;
    fact[0] = 1; 
    for(int i = 1; i < 9; i++) fact[i] = fact[i - 1] * i; 
    for(int i = 0; i < n; i++){
        int a; cin >> a;
        xd.push_back(a);
    }
    resp += posi();
    xd.clear();
    for(int i = 0; i < n; i++){
        int a; cin >> a;
        xd.push_back(a);
    }
    resp -= posi();
    if(resp < 0) resp *= -1;
    cout << resp << endl;

    return 0;


}