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 pair<int, int> ii;
typedef pair<int, ii> iii;
typedef vector<ii> vii;
typedef vector<int> vi;
typedef long long ll;
typedef long double ld;
typedef vector<ll> vll;
typedef pair<ll,ll> lll;

const ll MOD = 1000000007;
const ll INF = 1e9+5;
const double eps = 1e-7;
const double PI = acos(-1.0);

#define FOR(i,a,b) for(long long i = (long long)(a); i < (long long)(b); i++)
#define RFOR(i,a,b) for(long long i = (long long)(a); i >= (long long)(b); i--)
#define ull unsigned long long
#define deb(x )     cerr << #x << " = "<< x << endl;
#define endl    "\n"
#define coud(a,d) cout << fixed << showpoint << setprecision(d) << a;
#define ff first
#define ss second
#define mp make_pair
#define pb push_back
#define fill(x, y) memset(x, y, sizeof(y))
#define all(x) (x).begin(), (x).end()
#define bitcount(x) __builtin_popcount(x)
#define cps CLOCKS_PER_SEC
#define verify(x) if(x >= MOD) x -= MOD;

inline void debug_vi(vi a) {FOR(i, 0, a.size()) cout<<a[i]<<" ";}
inline void debug_vll(vll a) {FOR(i, 0, a.size()) cout<<a[i]<<" ";}
inline void print_case(int tn) {cout<<"Case #"<<tn<<": ";}

template<typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;

template<typename T>
using maxpq = priority_queue<T>;
/*----------------------------------------------------------------------*/

bool g(vi &a, vi &b) {
    FOR(i, 0, a.size()) {
        if(a[i] > b[i]) return true;
        if(a[i] < b[i]) return false;
    }
    return false;
}

bool eq(vi &a, vi &b) {
    FOR(i, 0, a.size()) {
        if(a[i] != b[i]) return false;
    }
    return true;
}

int main(){
    std::ios::sync_with_stdio(false);cin.tie(NULL); cout.tie(NULL);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    clock_t clk = clock();
    int n; cin>>n;
    vi a(n), b(n);
    FOR(i, 0, n) cin>>a[i];
    FOR(i, 0, n) cin>>b[i];
    if(g(a, b)) swap(a, b);
    int res = 0;
    while(true) {
        if(eq(a, b)) break;
        next_permutation(all(a));
        res++;
    }
    cout<<res;
    cerr << endl << "Time (in ms): " << double(clock() - clk) * 1000.0 / cps << endl;
    return 0;
}