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>
using namespace std;
#define int long long
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
using P=pair<long long,long long>;
#define rep(i,n) for(long long i=0; i<(long long)n; i++)
#define FOR(i,a,b) for(long long i=a; i<b; i++)
#define ll long long
#define all(x) (x).begin(), (x).end()
#define SZ(x) ((long long)(x).size())
#define COUT(x) cout << x << endl
#define PB(x) push_back(x)
#define MP make_pair
#define f first
#define s second
#define ld long double
#define vint vector<int>
#define vvint vector<vector<int>>
#define vstr vector<string>
#define vp vector<pair<int,int>>
#define vb vector<bool>
#define SUM(x) accumulate(x.begin(),x.end(),0)
#define MAX(x) *max_element(x.begin(),x.end())
#define MIN(x) *min_element(x.begin(),x.end())
#define couty cout<<'Y'<<'e'<<'s'<<endl
#define coutn cout<<'N'<<'o'<<endl
#define coutY cout<<'Y'<<'E'<<'S'<<endl
#define coutN cout<<'N'<<'O'<<endl
#define yn(x) cout<<(x?"Yes":"No")<<endl
#define YN(x) cout<<(x?"YES":"NO")<<endl
long long gcd(long long a,long long b){return b?gcd(b,a%b):a;}
long long lcm(long long a, long long b){return a*b/gcd(a,b);}
const long long dx[4]={1,0,-1,0};
const long long dy[4]={0,1,0,-1};
const long long INF = 1e12;
const long long MOD = 1e9+7;

signed main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout<<fixed<<setprecision(15);

    int n; cin >> n;
    vint p(n),q(n);
    rep(i,n) cin >> p[i];
    rep(i,n) cin >> q[i];
    vint order(n);
    rep(i,n) order[i]=i+1;
    int res=0;
    int count=0;
    do{
        count++;
        if(order==p) res+=count;
        if(order==q) res-=count;
    }while(next_permutation(all(order)));
    cout<<abs(res)<<endl;
}