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;
const static ll INF = 1e15;
// __uint128_t



int main(){
    ll N; cin >> N;
    vector<ll> A(N), B(N);
    for(int i = 0; i < N; i++) cin >> A[i];
    for(int i = 0; i < N; i++) cin >> B[i];

    vector<vector<ll>> DP((1<<N)+1, vector<ll>(55, INF)); DP[0][0] = 0;
    for(int i = 0; i < (1<<N) ; i++){
        bitset<20> b = i;
        ll c = b.count();


        ll p = 0; // 立っているビットが左にソートされた上で、自分がどの位置にいるか
        for(int j = N-1; j >= 0; j--){
            if(b[j] == 1) {p++; continue;}
            for(int k = 0; k <= 50; k++){
                if(DP[i][k] == INF) continue;
                bitset<20> b2 = 0; b2[j] = 1;
                ll n;
                if( abs(c-j)%2 == 0) n = A[j];
                else n = B[j];
                if(n >= k) { //照準にならないなら棄却
                    DP[(b|b2).to_ullong()][n] = min(DP[i][k] + abs(c-(j+p)), DP[(b|b2).to_ullong()][n]);
                }
            }
        }
    }
    ll ans = INF;
    for(int i = 0; i <= 50; i++) ans = min(ans, DP[(1<<N)-1][i]);
    if(ans == INF) cout << -1 << endl;
    else cout << ans << endl;



}