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>

bool vector_eq(const std::vector<int> &a, const std::vector<int> &b) {
    for(uint32_t i = 0; i < a.size(); ++i) {
        if(a[i] != b[i]) {
            return false;
        }
    }
    return true;
}

int main() {
    int n;
    std::cin >> n;

    auto ps = std::vector<int>();
    ps.reserve(n);
    for(int i = 0; i < n; ++i) {
        int p;
        std::cin >> p;
        ps.push_back(p);
    }

    auto qs = std::vector<int>();
    qs.reserve(n);
    for(int i = 0; i < n; ++i) {
        int q;
        std::cin >> q;
        qs.push_back(q);
    }

    auto v = std::vector<int>();
    v.reserve(n);
    for(int i = 1; i <=n; ++i) {
        v.push_back(i);
    }

    int64_t p_pos, q_pos;
    int64_t pos = 0;
    int found = 0;

    do {
        ++pos;
        if(vector_eq(ps, v)) {
            p_pos = pos;
            ++found;
        }
        if(vector_eq(qs, v)) {
            q_pos = pos;
            ++found;
        }
        if(found >= 2) {
            break;
        }
    } while(std::next_permutation(v.begin(), v.end()));

    std::cout << std::abs(p_pos - q_pos);

    return 0;
}