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;

int n;
int a[20], b[20];
int ans = 1e6;

int DP[1<<18][18];

int getVal(int x, int y){
    if((x+y)&1) return a[y];
    return b[y];
}

int main(){
    scanf("%d", &n);
    for(int i=0; i<n; i++){
        scanf("%d", &a[i]);
    }
    for(int i=0; i<n; i++){
        scanf("%d", &b[i]);
    }

    for(int i=1; i<(1<<n); i++){
        int pcnt = __builtin_popcount(i);
        if(pcnt == 1){
            for(int j=0; j<n; j++){
                if(i&(1<<j)) DP[i][j] = 0;
                else DP[i][j] = 1e6;
            }
            continue;
        }
        for(int j=0; j<n; j++){
            DP[i][j] = 1e6;
            if(!(i&(1<<j))) continue;
            int addInv = 0;
            for(int k=0; k<n; k++){
                if(!(i&(1<<k))) continue;
                if(j < k) addInv++;
            }

            for(int k=0; k<n; k++){
                if(!(i&(1<<k))) continue;
                if(getVal(pcnt, j) >= getVal(pcnt-1, k)){
                    DP[i][j] = min(DP[i][j], DP[(i-(1<<j))][k] + addInv);
                }
            }
        }
    }
    for(int i=0; i<n; i++) ans = min(ans, DP[(1<<n)-1][i]);
    printf("%d", ans==1e6?-1:ans);
}