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;

#define rep(i, n) for(int i = 0; i < (int)n; ++i)
#define FOR(i, a, b) for(int i = a; i < (int)b; ++i)
#define rrep(i, n) for(int i = ((int)n - 1); i >= 0; --i)

typedef long long ll;
typedef long double ld;

const ll Inf = 1e18;
const double EPS = 1e-9;
const int MOD = 1e9 + 7;

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(0);
    int n, cnt = 1;
    cin >> n;
    vector<int> p(n), q(n), num(n);
    rep (i, n) {
        cin >> p[i];
        num[i] = i + 1;
    }
    rep (i, n) cin >> q[i];
    int a = -1, b = -1;
    do {
        if (a == -1) {
            bool ok = true;
            rep (i, n) if (num[i] != p[i]) ok = false; 
            if (ok) a = cnt;
        }
        if (b == -1) {
            bool ok = true;
            rep (i, n) if (num[i] != q[i]) ok = false;
            if (ok) b = cnt;
        }
        cnt++;
    } while (next_permutation(num.begin(), num.end()));
    cout << abs(a - b) << endl;
    
    return 0;
}