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++ (GCC 9.2.1) (judge verdict: Accepted)

using namespace std;
#include <bits/stdc++.h>
#define ll long long
#define ld long double
#define EL '\n'
#define el '\n'
#define all(x) x.begin(),x.end()
#define rep(i, a, b) for (ll i = (a); i < (b); i++)
#define repd(i, a, b) for (ll i = (b)-1; i >= (a); i--)
#define large 1000000007
#define MOD 1000000007
#define pb push_back
#define MAX 10e+7
#define FASTIO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
void solve()
{
int n;
cin>>n;
string s1,s2;
for (int i = 0; i < n; i++)
{
    int a;
    cin>>a;
    s1+=a;
}
for (int i = 0; i < n; i++)
{
    int a;
    cin>>a;
    s2+=a;
}
string s11=s1,s22=s2;
sort(s1.begin(),s1.end());
sort(s2.begin(),s2.end());
int ans1=0,ans2=0;
do
{
    ans1++;
    if(s1==s11)
    goto end;
} while (next_permutation(s1.begin(),s1.end()));
end:
do
{
ans2++;
if(s2==s22)
goto end1;
} while (next_permutation(s2.begin(),s2.end()));
end1:
cout<<abs(ans1-ans2)<<endl;
}
int main()
{

FASTIO
 long long int t;
 t=1;
 while(t--)
 {
  solve();
  
 }
  return 0;
}