Problem: 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.

Examples

Input

3
3 4 3
3 2 3


Output

1


Input

2
2 1
1 2


Output

-1


Input

4
1 2 3 4
5 6 7 8


Output

0


Input

5
28 15 22 43 31
20 22 43 33 32


Output

-1


Input

5
4 46 6 38 43
33 15 18 27 37


Output

3

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
#define rep(i,n) for (ll i = 0; i < (n); ++i)
#define vcout(v) rep(i,v.size()) cout << v[i] << " "; cout << endl
#define mcout(m) rep(j,m.size()){vcout(m[j]);}
const ll MOD = 1e9 + 7;
const ll INF = 1e18;
void mainsolve(){
	ll n;
	cin >> n;
	vector<ll> a(n), b(n);
	rep(i,n) cin >> a[i];
	rep(i,n) cin >> b[i];

	vector<vector<ll>> dp(1<<n, vector<ll>(55, INF));
	dp[0][0] = 0;
	rep(i,1<<n){
		ll cnt = __builtin_popcount(i);
      ll cnt2 = cnt;
		vector<P> ords;
		rep(k,n){
			if(!(i & (1<<k))){
				ords.push_back(P(k,cnt2++));
			}
		}

		rep(s,55){
			for(auto p : ords){
				ll t;
				if(abs(p.first-cnt) % 2 == 0) t = a[p.first];
				else t = b[p.first];
				if(t >= s){
					ll next = i | (1<<p.first);
					dp[next][t] = min(dp[next][t], dp[i][s] + abs(p.second-cnt));
				}
			}
		}
	}

	ll res = INF;
	rep(s,55){
		res = min(res, dp[(1<<n) - 1][s]);
	}
	if(res == INF) cout << -1 << endl;
	else cout << res << endl; // res = 全カードの移動距離
	return;
}

int main(){
	
	//ll TC;cin>>TC;rep(I,TC){cout<<"case:"<<I+1<<"---"<<endl; // comment out!
		mainsolve();
	//} // comment out!
	return 0;
}