Problem:
Score : 700 points

Problem Statement

We have N cards numbered 1, 2, ..., N.
Card i (1 \leq i \leq N) has an integer A_i written in red ink on one side and an integer B_i written in blue ink on the other side.
Initially, these cards are arranged from left to right in the order from Card 1 to Card N, with the red numbers facing up.

Determine whether it is possible to have a non-decreasing sequence facing up from left to right (that is, for each i (1 \leq i \leq N - 1), the integer facing up on the (i+1)-th card from the left is not less than the integer facing up on the i-th card from the left) by repeating the operation below. If the answer is yes, find the minimum number of operations required to achieve it.

Choose an integer i (1 \leq i \leq N - 1).
Swap the i-th and (i+1)-th cards from the left, then flip these two cards.

Constraints

1 \leq N \leq 18

1 \leq A_i, B_i \leq 50 (1 \leq i \leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N
B_1 B_2 ... B_N

Output

If it is impossible to have a non-decreasing sequence, print -1.
If it is possible, print the minimum number of operations required to achieve it.

Sample Input 1

3
3 4 3
3 2 3

Sample Output 1

1

By doing the operation once with i = 1, we have a sequence [2, 3, 3] facing up, which is non-decreasing.

Sample Input 2

2
2 1
1 2

Sample Output 2

-1

After any number of operations, we have the sequence [2, 1] facing up, which is not non-decreasing.

Sample Input 3

4
1 2 3 4
5 6 7 8

Sample Output 3

0

No operation may be required.

Sample Input 4

5
28 15 22 43 31
20 22 43 33 32

Sample Output 4

-1

Sample Input 5

5
4 46 6 38 43
33 15 18 27 37

Sample Output 5

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
constexpr int N = 18;
int n;
int a[N][2];
int dp[1 << N][N];
int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cin >> n;
    for (int i = 0; i < n; ++i)
        std::cin >> a[i][0];
    for (int i = 0; i < n; ++i) {
        std::cin >> a[i][1];
        if (i % 2 == 1)
            std::swap(a[i][0], a[i][1]);
    }
    for (int i = 0; i < (1 << n); ++i)
        std::fill(dp[i], dp[i] + n, -1);
    for (int i = 0; i < n; ++i)
        dp[1 << i][i] = 0;
    for (int s = 1; s < (1 << n) - 1; ++s) {
        int pos = __builtin_popcount(s) - 1;
        for (int i = 0; i < n; ++i) {
            if ((s >> i & 1) && dp[s][i] != -1) {
                int val = a[i][pos % 2];
                for (int j = 0; j < n; ++j) {
                    if ((~s >> j & 1) && a[j][(pos + 1) % 2] >= val) {
                        int t = s | 1 << j;
                        int tmp = dp[s][i] + pos + 1 - __builtin_popcount(s & ((1 << j) - 1));
                        if (dp[t][j] == -1 || dp[t][j] > tmp)
                            dp[t][j] = tmp;
                    }
                }
            }
        }
    }
    int ans = -1, s = (1 << n) - 1;
    for (int i = 0; i < n; ++i)
        if (dp[s][i] != -1 && (ans == -1 || ans > dp[s][i]))
            ans = dp[s][i];
    std::cout << ans << "\n";
    return 0;
}