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 "bits/stdc++.h"
using namespace std;
typedef long long ll;


#include "boost/multiprecision/cpp_int.hpp"
#include "boost/multiprecision/cpp_dec_float.hpp"
namespace mp = boost::multiprecision;
typedef mp::cpp_int LL;
typedef mp::number<mp::cpp_dec_float<1024>> DD;// 仮数部が1024ビットの浮動小数点数型(TLEしたら小さくする)

 
typedef long double dd;
#define i_7 (ll)(1E9+7)
//#define i_7 998244353
#define i_5 i_7-2
ll mod(ll a){
    ll c=a%i_7;
    if(c>=0)return c;
    return c+i_7;
}
typedef pair<ll,ll> l_l;
typedef pair<dd,dd> d_d;
ll inf=(ll)1E16;
#define rep(i,l,r) for(ll i=l;i<=r;i++)
#define rrep(i,r,l) for(ll i=r;i>=l;i--)
#define pb push_back
ll max(ll a,ll b){if(a<b)return b;else return a;}
ll min(ll a,ll b){if(a>b)return b;else return a;}
void Max(ll &pos,ll val){pos=max(pos,val);}//Max(dp[n],dp[n-1]);
void Min(ll &pos,ll val){pos=min(pos,val);}
void Add(ll &pos,ll val){pos=mod(pos+val);}
dd EPS=1E-9;
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define fi first
#define se second
#define endl "\n"
#define SORT(v) sort(v.begin(),v.end())
#define ERASE(v) v.erase(unique(v.begin(),v.end()),v.end())
#define POSL(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define POSU(v,x) (upper_bound(v.begin(),v.end(),x)-v.begin())
template<class T,class S>
inline bool chmax(T &a, S b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}
template<class T,class S>
inline bool chmin(T &a, S b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

void mod_print(ll k){
    ll P=1000;
    rep(y,1,P){
        ll x=mod(y*k);
        if(x+P>=i_7){
            x-=i_7;
        }
        if(abs(x)<=P){
            cout<<x<<"/"<<y<<endl;
            return;
        }
    }
    cout<<"nun"<<endl;
}
#define all(c) c.begin(),c.end()
//////////////////////////


int main(){fastio
    ll n;cin>>n;
    ll p[n],q[n];
    
    rep(i,0,n-1)cin>>p[i];
    rep(i,0,n-1)cin>>q[i];
    ll a[n];rep(i,0,n-1)a[i]=i+1;
    ll c=0,a1,a2;
    do{
        c++;
        bool f=true;
        rep(i,0,n-1){
            if(p[i]!=a[i])f=false;
        }
        if(f)a1=c;
        f=true;
        rep(i,0,n-1){
            if(q[i]!=a[i])f=false;
        }
        if(f)a2=c;
    }while(next_permutation(a,a+n));
    
    cout<<abs(a2-a1)<<endl;
    
    return 0;
}