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>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007, INF=mod*mod*3LL;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;

void chmin(int &a, int b){a = min(a, b);}
int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    int n, a = 1e9;
    cin >> n;
    vector<vector<int> > v(n, vector<int> (2));
    for(int i=0;i<n;i++) cin >> v[i][0];
    for(int i=0;i<n;i++) cin >> v[i][1];
    int maki = 1 << n;
    vector<vector<int> > dp(maki, vector<int> (51, a));
    dp[0][0] = 0;
    for(int bit=0;bit<maki;bit++){
    	int cnt = 0;
    	vector<int> d(n);
    	for(int i=0;i<n;i++) if(!(bit>>i&1)) d[i] = cnt++;
    	for(int i=0;i<n;i++){
    		int k = abs(i - n + cnt);
    		for(int j=0;j<=v[i][k&1];j++)
    			chmin(dp[bit|(1<<i)][v[i][k&1]], dp[bit][j] + d[i]);
    	}
    }
    for(int i=0;i<=50;i++) chmin(a, dp[maki-1][i]);
    cout << (a==1e9? -1: a) << endl;
    return 0;
}