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;

using lint = long long;
using pint = pair<int, int>;
using plint = pair<lint, lint>;

constexpr lint INF = 1'010'000'000'000'000'017LL;
constexpr lint MOD = 1'000'000'007LL;
constexpr double EPS = 1e-12;
constexpr double PI = 3.14159265358979323846;

#define FOR(i, start, end) for(lint i = (start); i < end; i++)
#define REP(i, n) FOR(i, 0, n)
#define ALL(v) v.begin(), v.end()

template<class T> inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }

lint mod(lint val, lint m){ lint res = val % m; if(res < 0) res += m; return res;}
lint modinv(lint a, lint m) { lint b = m, u = 1, v = 0; while (b) { lint t = a / b; a -= t * b; swap(a, b); u -= t * v; swap(u, v); } u %= m; if (u < 0) u += m; return u;}
lint modpow(lint a, lint n, lint mod) { lint res = 1; while (n > 0) { if (n & 1) res = res * a % mod; a = a * a % mod; n >>= 1;} return res;}
lint gcd(lint a, lint b){ if(a < b) swap(a, b); if(b == 0) return a; return gcd(b, a%b);}
lint lcm(lint a, lint b){ return a * b / gcd(a, b);}


//---------------------------------------------------------------

void Main()
{
    int N, a, b, ans;
    cin >> N;
    vector<int> P(N), Q(N), nums(N);

    REP(i, N){
        nums[i] = i+1;
        cin >> P[i];
    }
    REP(i, N){
        cin >> Q[i];
    }

    a = -1;
    b = -1;
    int i = 0;
    do {
        i++;
        if(P == nums){
            a = i;
        }
        if(Q == nums){
            b = i;
        }
        if(a != -1 && b != -1){
            break;
        }
    } while(next_permutation(nums.begin(), nums.end()));

    ans = abs(a-b);
    cout << ans << endl;
}

//---------------------------------------------------------------
int main()
{
    cout << fixed << setprecision(15);
    Main();
    return 0;
}