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:
//https://atcoder.jp/contests/keyence2020/tasks/keyence2020_d

#include<bits/stdc++.h>
const int inf = 1e9;
using namespace std;

int n, ans = inf, a[18], b[18], dp[1 << 18][18];

int val(int i, int j){
    j &= 1;
    return (j ? b[i] : a[i]);
}

void up(int&a, int b){
    if (a > b) a = b;
}

int main(){
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> n;

    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) cin >> b[i];
    for (int i = 0; i < n; i++) if (i&1) swap(a[i], b[i]);

    for (int i = 0; i < (1 << n); i++) for (int j = 0; j < n; j++) dp[i][j] = inf;
    for (int i = 0; i < n; i++) dp[1 << i][i] = i;

    for (int i = 1; i < (1 << n); i++) if (i & (i - 1)){
        int cnt = 0;
        for (int j = 0; j < n; j++) {
            if (((i >> j)&1)){
                int num = __builtin_popcount(i);
                for (int k = 0; k < n; k++) if (k != j && val(k, num-2) <= val(j, num-1)){
                    up(dp[i][j], dp[i^(1 << j)][k] + cnt);
                }
            }
            else cnt++;
        }
    }
    for (int i = 0; i < n; i++) up(ans, dp[(1 << n)-1][i]);
    if (ans == inf) ans = -1;
    cout << ans;
}