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;
#define rep(i,n) for(int i = 0; i < (n); i++)
#define ALL(v) (v).begin(),(v).end()
using ll = long long;
using P = pair<int, int>;
const int INF = 1e9;
const long long LINF = 1e18;
const long long MOD = 1e9 + 7;

#define MAX 50

signed main(){
    int n;
    cin >> n;
    int c[2][n];
    rep(i,n){
        cin >> c[0][i];
    }
    rep(i,n){
        cin >> c[1][i];
    }
    vector<vector<int>> dp(1 << n, vector<int>(MAX + 1, INF));
    dp[0][0] = 0;
    for(int bit = 0; bit < 1 << n; bit++){
        int now = 0;
        int dist[n];
        for(int i = 0; i < n; i++){
            if(!(bit & 1 << i)){
                dist[i] = now;
                now++;
            }
        }
        int cnt = n - now;
        for(int i = 0; i < n; i++){
            if(bit & (1 << i)) continue;
            int k = abs(i - cnt);
            for(int j = 0; j <= c[k % 2][i]; j++){
                dp[bit | (1 << i)][c[k % 2][i]] = min(dp[bit | (1 << i)][c[k % 2][i]], dp[bit][j] + dist[i]);
            }
        }
    }
    int ans = INF;
    for(int i = 0; i <= 50; i++){
        ans = min(ans, dp[(1 << n) - 1][i]);
    }
    cout << (ans == INF ? -1 : ans) << endl;
    return 0;
}