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;


/// File input & output /////
#define fRead(x)        freopen("x.txt","r",stdin)
#define fWrite(x)       freopen ("x.txt","w",stdout)


#define ld              long double
#define ll              long long
#define ull             unsigned long long
#define ff              first
#define ss              second
#define pb              push_back
#define INF             2e16
#define PI              acos(-1.0)
#define mp              make_pair
#define pii             pair<int,int>
#define pll             pair<LL,LL>


#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(a,min(b,min(c,d)))
#define max4(a,b,c,d)   max(a,max(b,max(c,d)))
#define SQR(a)          ((a)*(a))
#define FOR(i,a,b)      for(int i=a;i<=b;i++)
#define ROF(i,a,b)      for(int i=a;i>=b;i--)
#define REP(i,b)        for(int i=0;i<b;i++)
#define MEM(a,x)        memset(a,x,sizeof(a))
#define ABS(x)          ((x)<0?-(x):(x))

#define SORT(v)         sort(v.begin(),v.end())
#define REV(v)          reverse(v.begin(),v.end())


#define FASTIO        ios_base::sync_with_stdio(0);cin.tie(nullptr);




///..................Sort by second element in a pair in decending order............. ///
bool SortbySecDesc(const pair<long long,long long > &a, const pair<long long,long long > &b)
{
    return a.second>b.second;
}

///.........................first element of pair in descending order..........................//////
bool sortinrev(const pair<int,int> &a,
               const pair<int,int> &b)
{
       return (a.first > b.first);
}



///.............................................................TEMPLATE................................................./////

int main()
{
    FASTIO;

    int n;

    cin >> n;

    string s1 , s2;



    for(int i=0; i<n; i++)
    {
        int a;

        cin >> a;

        s1 = s1 + to_string(a);
    }

    for(int i=0; i<n; i++)
    {
        int b;

        cin >> b;

        s2 += to_string(b);
    }

    if(s1==s2)
    {
        cout << 0 << endl;
        return 0;
    }



    string ans;


    for(int i=1; i<=n; i++)
    {
        ans += to_string(i);
    }

   // cout << s1 << " " << s2 << endl;


      int k = 0;

      int pos , pos1 ;


     do{

           // cout << ans << endl;
            k++;

        if(ans==s1)
        {
            pos = k;
        }
        if(ans==s2)
        {
            pos1 = k;
        }

     } while(next_permutation(ans.begin(),ans.end()));

    // cout << pos << " " << pos1 << endl;


      cout << abs(pos-pos1) << endl;



    return 0;
}