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 int long long

int memo[20][(1 << 18)];
int n;
int a[20];
int b[20];

int get(int i, int j){
    if(abs(j-i)%2 == 0){
        return a[j];
    }else{
        return b[j];
    }
}

int dp(int i, int bit, int last, int num){
    if(i == n){
        return 0;
    }

    if(memo[last][bit] != -1) return memo[last][bit];

    int ans = INT_MAX;
    int cnt = 0;
    for(int j = 0;j < n;j++){
        if((bit & (1 << j)) == 0){
            if(num <= get(i, j)){
                ans = min(ans, dp(i+1, bit | (1 << j), j, get(i, j)) + j-cnt);
            }
        }else{
            cnt++;
        }
    }

    return memo[last][bit] = ans;
}

signed main(){
    cin >> n;

    for(int i = 0;i < n;i++) cin >> a[i];
    for(int i = 0;i < n;i++) cin >> b[i];

    memset(memo, -1, sizeof(memo));

    cout << (dp(0, 0, n, 0) == INT_MAX ? -1 : dp(0, 0, n, 0)) << endl;
    return 0;
}