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 int ll;
typedef long double ld;
typedef vector<ll> vl;
typedef vector<string> vs;
typedef vector<char> vc;
typedef queue<ll> ql;
typedef deque<ll> dql;
typedef priority_queue<ll, vl/*, greater<ll>*/> pql; //降順(/*昇順*/)
typedef set<ll> sl;
typedef pair<ll, ll> pl;
typedef vector<vl> vvl;
typedef vector<pl> vpl;
#define rep(i, n) for(ll i = 0; i < ll(n); ++i)
#define rep2(i, n) for(ll i = 1; i <= ll(n); ++i)
//#define rep(i, k, n) for(ll i = k-1; i < ll(n); ++i)
//#define rep2(i, k, n) for(ll i = k; i <= ll(n); ++i)
#define all(v) (v).begin(), (v).end()
bool chmin(ll &a, ll b) {if(b < a) {a = b; return 1;} return 0;}
bool chmax(ll &a, ll b) {if(b > a) {a = b; return 1;} return 0;}
const ll INF = 1LL << 60;
const ll MOD = 1e9 + 7;
//const ll MOD = 998244353;
const ll MAX = 1e9;
const char newl = '\n';

void input(string &S) {
  char c;
  cin >> c;
  S+=c;
  return;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  ll n;
  cin >> n;
  string p, q;
  char c;
  rep(i, n) input(p);
  rep(i, n) input(q);
  if(p>q) swap(p, q);
  vl P(n), Q(n);
  rep(i, n) P[i] = p[i]-'0';
  rep(i, n) Q[i] = q[i]-'0';
  ll ans=0;
  do {
    if(P==Q) break;
    ans++;
  } while (next_permutation(all(P)));
  cout << ans << newl;
  return 0;
}