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 ll = long long;

const int MOD = 1e9 + 7;
#define REP(i, n)  for(int i = 0; i < n; ++i)
#define FOR(i, t, n)  for(ll i = t; i <= (ll)n; ++i)
struct edge {
    int from, to;
    ll cost;
};


class Eratosthenes {
public:
    vector<bool> prime;

    Eratosthenes(ll size) {
        size = max((int) sqrt(size) + 1, 101010);
        prime.resize(size, 1);
        prime[0] = 0;
        prime[1] = 0;
        for (size_t i = 2; i * i <= size; ++i)
            if (prime[i]) {
                for (size_t j = 2; i * j <= size; ++j)prime[i * j] = false;
            }
    }
};


class Combination {
    vector<ll> fac; //n!(mod M)
    vector<ll> ifac; //k!^{M-2} (mod M)
public:
    Combination(ll size) {
        fac.resize(size);
        ifac.resize(size);
        fac[0] = 1;
        ifac[0] = 1;
        for (ll i = 0; i < size; i++) {
            fac[i + 1] = fac[i] * (i + 1) % MOD;
            ifac[i + 1] = ifac[i] * mpow(i + 1, MOD - 2) % MOD;
        }
    }

    ll mpow(ll x, ll n) {
        ll ans = 1;
        while (n != 0) {
            if (n & 1) ans = ans * x % MOD;
            x = x * x % MOD;
            n = n >> 1;
        }
        return ans;
    }

    ll comb(ll a, ll b) {
        if (a == 0 && b == 0)return 1;
        if (a < b || a < 0)return 0;
        ll tmp = ifac[a - b] * ifac[b] % MOD;
        return tmp * fac[a] % MOD;
    }
};


ll GCD(ll a, ll b) {
    if (b == 0)return a;
    return GCD(b, a % b);
}

ll LCM(ll a, ll b) {
    ll gcd = GCD(a, b);
    return a / gcd * b;
}

ll t1, t2, a1, a2, b1, b2;
string s;
ll N;

int main() {
    cin >> N;
    vector<int> A(N), B(N), x, y;
    REP(i, N)cin >> A[i];
    REP(i, N)cin >> B[i];
    x = A,y = B;
    sort(A.begin(), A.end());
    sort(B.begin(), B.end());
    ll a, b, cnt = 1;
    do {
        bool isTrue = true;
        REP(i, N)if (A[i] != x[i])isTrue = false;
        if (isTrue)a = cnt;
        cnt++;
    } while (next_permutation(A.begin(), A.end()));
    cnt = 1;
    do {
        bool isTrue = true;
        REP(i, N)if (B[i] != y[i])isTrue = false;
        if (isTrue)b = cnt;
        cnt++;
    } while (next_permutation(B.begin(), B.end()));
    cout << abs(a - b) << endl;
    return 0;
}