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 <bits/stdc++.h>
using namespace std;
using LL = long long int;
#define incID(i, l, r) for(int i = (l)    ; i <  (r); ++i)
#define incII(i, l, r) for(int i = (l)    ; i <= (r); ++i)
#define decID(i, l, r) for(int i = (r) - 1; i >= (l); --i)
#define decII(i, l, r) for(int i = (r)    ; i >= (l); --i)
#define inc(i, n)  incID(i, 0, n)
#define inc1(i, n) incII(i, 1, n)
#define dec(i, n)  decID(i, 0, n)
#define dec1(i, n) decII(i, 1, n)
#define inID(v, l, r) ((l) <= (v) && (v) <  (r))
#define inII(v, l, r) ((l) <= (v) && (v) <= (r))
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define FI first
#define SE second
#define  ALL(v)  v.begin(),  v.end()
#define RALL(v) v.rbegin(), v.rend()
auto setmin   = [](auto & a, auto b) { return (b <  a ? a = b, true : false); };
auto setmax   = [](auto & a, auto b) { return (b >  a ? a = b, true : false); };
auto setmineq = [](auto & a, auto b) { return (b <= a ? a = b, true : false); };
auto setmaxeq = [](auto & a, auto b) { return (b >= a ? a = b, true : false); };
auto fl = [](auto a, auto b) { assert(b != 0); return a / b - (a % b != 0 && ((a >= 0) != (b >= 0)) ? 1 : 0); };
auto ce = [](auto a, auto b) { assert(b != 0); return a / b + (a % b != 0 && ((a >= 0) == (b >= 0)) ? 1 : 0); };
auto mo = [](auto a, auto b) { assert(b != 0); a %= b; if(a < 0) { a += abs(b); } return a; };
LL gcd(LL a, LL b) { return (b == 0 ? a : gcd(b, a % b)); }
LL lcm(LL a, LL b) { return a / gcd(a, b) * b; }
#define bit(b, i) (((b) >> (i)) & 1)
#define SI(v) static_cast<int>(v.size())
#define RF(e, v) for(auto & e: v)
#define until(e) while(! (e))
#define if_not(e) if(! (e))
#define ef else if
#define UR assert(false)

// ---- ----

int main() {
	int n;
	cin >> n;
	vector<int> a(n), b(n);
	inc(i, n) { cin >> a[i]; }
	inc(i, n) { cin >> b[i]; }
	
	int INF = 1e9, ans = INF;
	inc(B, 1 << n) {
		vector<pair<int, int>> v, x, y, w;
		inc(i, n) { v.EB((bit(B, i) == 0 ? a[i] : b[i]), i); }
		
		inc(i, n) { (bit(B, i) == i % 2 ? x : y).PB(v[i]); }
		if_not(SI(x) == SI(y) || SI(x) == SI(y) + 1) { continue; }
		
		sort(ALL(x));
		sort(ALL(y));
		inc(i, SI(x)) {
			w.PB(x[i]);
			if(i < SI(y)) { w.PB(y[i]); }
		}
		bool f = true;
		inc(i, n - 1) { f &= (w[i].FI <= w[i + 1].FI); }
		if(! f) { continue; }
		
		int t = 0;
		inc(i, n) {
		inc(j, n) {
			if(i < j && w[i].SE > w[j].SE) { t++; }
		}
		}
		setmin(ans, t);
	}
	
	if(ans == INF) { ans = -1; }
	cout << ans << endl;
	
	return 0;
}