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>
#include <unistd.h>
 
using namespace std;
 
typedef long long ll;
typedef pair<ll,ll> P;
 
long long int INF = 3e18;
const ll fact_table = 800008;
double Pi = 3.1415926535897932384626;
 
//vector<ll> G[550010];
vector<P> tree[500010];
priority_queue <ll> pql;
priority_queue <P> pqp;
//big priority queue
priority_queue <ll,vector<ll>,greater<ll> > pqls;
priority_queue <P,vector<P>,greater<P> > pqps;
//small priority queue
//top pop
 
ll dx[8]={1,0,-1,0,1,1,-1,-1};
ll dy[8]={0,1,0,-1,1,-1,-1,1};
char dir[] = "DRUL";
//ll bit[500005];
//↓,→,↑,←
 
 
#define p(x) cout<<x<<"\n";
#define el cout<<endl;
#define pe(x) cout<<(x)<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<(x);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define CLEAR(a) a = decltype(a)(); 
 
 
//ll mod = 998244353;
ll mod = 1000000007;
 
ll rui(ll number1,ll number2){
 
    if(number2 == 0){
        return 1;
    }else{
        ll number3 = rui(number1,number2 / 2);
        number3 *= number3;
        number3 %= mod;
        if(number2%2==1){
            number3 *= number1;
            number3 %= mod;
        }
        return number3;
    }
}
ll gcd(ll number1,ll number2){
 
    if(number1 > number2){
        swap(number1,number2);
    }
    if(number1 == 0 || number1 == number2){
        return number2;
    }else{
        return gcd(number2 % number1,number1);
    }
}
void YES(bool condition){
 
    if(condition){
        p("YES");
    }else{
        p("NO");
    }
    return;
}
void Yes(bool condition){
 
    if(condition){
        p("Yes");
    }else{
        p("No");
    }
    return;
}
 
/*
ll fact[fact_table + 5],rfact[fact_table + 5];
 
void c3_init(){
    fact[0] = rfact[0] = 1;
    for(ll i=1; i<=fact_table; i++){
        fact[i] = (fact[i-1]*i) % mod;
    }
    rfact[fact_table] = rui(fact[fact_table],mod - 2);
    for(ll i=fact_table; i>=1; i--){
       rfact[i-1] = rfact[i] * i;
       rfact[i-1] %= mod;
    }
    return;}
ll c3(ll n,ll r){
    return (((fact[n] * rfact[r]) % mod ) * rfact[n-r]) % mod;}
*/
 
ll n,m,num,sum,a,b,c,d,e,g,h,w,i,j,q,r,l;
ll k;
ll ans;
ll x[200005], y[500005], z[200005];
bool used[500005];
char s[500005];
ll query;

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

    cin >> n;
    for(ll i=0;i<n;i++){
        cin >> y[i];
        x[i] = i + 1;
    }
    for(ll i=0;i<n;i++){
        cin >> z[i];
    }
    ll cnt = 0;
    do{
        bool is_y = true, is_z = true;
        for(ll i=0;i<n;i++){
            if(x[i] != y[i])is_y = false;
            if(x[i] != z[i])is_z = false;
        }
        if(is_y){
            a = cnt;
        }
        if(is_z){
            b = cnt;
        }
        cnt ++;
    }while(next_permutation(x, x + n));
    p(abs(a - b));


    return 0;
}