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 INF 10000

int N;
int A[20], B[20];
int dp[1<<20][20];

int main() {
    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<(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] = 0;
    
    for (int S=0; S<(1<<N); S++) {
        int cnt = 0;
        for (int i=0; i<N; i++) if ((S>>i)&1) cnt++;
        
        for (int i=0; i<N; i++) {
            int cost = cnt;
            
            for (int j=0; j<N; j++) {
                if (!((S>>j)&1)) {
                    int n1 = (i-(cnt-1))%2==0 ? A[i] : B[i];
                    int n2 = (j-cnt)%2==0 ? A[j] : B[j];
                    if (n1>n2) continue;
                    dp[S|(1<<j)][j] = min(dp[S|(1<<j)][j], dp[S][i]+cost);
                }
                else cost--;
            }
        }
    }
    
    int ans = INF;
    for (int i=0; i<N; i++) ans = min(ans, dp[(1<<N)-1][i]);
    cout << (ans==INF ? -1 : ans) << endl;
}