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;
typedef long long ll;

ll comb(ll N){
  if(N==0){
    return 0;
  }
  
  ll res=1;

  for(int i=0;i<N;i++){
    res *= (i+1);
  }

  return res;
}

int main(void)
{
  ll N;cin>>N;
  vector<ll> P(N);
  vector<ll> Q(N);

  for(int i=0;i<N;i++){
    cin>>P[i];
  }

  for(int i=0;i<N;i++){
    cin>>Q[i];
  }

  vector<int> Nums1(N);
  vector<int> Nums2(N);
  for(int i=0;i<N;i++){
    Nums1[i]=i+1;
    Nums2[i]=i+1;
  }

  ll pl1=0;
  ll pl2=0;
  for(ll i=0;i<N;i++){
    auto minnum = min_element(Nums1.begin(),Nums1.end());
    ll num = P[i]-*minnum;
    int times = find(Nums1.begin(),Nums1.end(),P[i])-Nums1.begin();
    //cout<<times<<"!"<<num<<"!"<<*minnum<<"!"<<pl1<<endl;
    pl1 += comb(N-i-1)*times;
    Nums1.erase(find(Nums1.begin(),Nums1.end(),P[i]));
  }

  for(ll i=0;i<N;i++){
    auto minnum = min_element(Nums2.begin(),Nums2.end());
    ll num = Q[i]-*minnum;
    int times = find(Nums2.begin(),Nums2.end(),Q[i])-Nums2.begin();
    pl2 += comb(N-i-1)*times;
    //cout<<num<<"!"<<*minnum<<"!"<<pl2<<endl;
    Nums2.erase(find(Nums2.begin(),Nums2.end(),Q[i]));
  }
  
  //cout<<pl2<<endl;

  ll res = abs(pl1-pl2);

  cout<<res<<endl;

  return 0;
}