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 <iostream>
#include <algorithm>
#include <vector>
#include <bitset>
using namespace std;
int n;
vector<int> a, b;

vector< int > dp;
int d2i(int bi, int usd){
    int id = bi;
    id *= (1<<n); id += usd;
    return id;
}

int f(int bi, int usd){
    if(usd == ((1<<n)-1)) return 0;
    int id = d2i(bi, usd);
    if(dp[id] != -1) return dp[id];
    int res = n*n, t = 0, s= 0;
    for(int i=0;i<n;i++) t += ((usd>>i)&1);
    for(int i=n-1;i>=0;i--){
        if(((usd>>i)&1) == 0){
            if(t > 0){
                int iv, bv;
                if((t+i)%2 == 0) iv = a[i];
                else iv = b[i];
                if((t+bi)%2 == 1) bv = a[bi];
                else bv = b[bi];

                if(iv < bv) continue;
            }
            int u = s;
            u += f(i, usd | (1<<i));
            if(res > u) res = u;
        }else{
            s++;
        }
    }

    //cerr << bi << " " << bitset<3>(usd) << " : " << res << endl;

    return dp[id] = res;
}

int main(void){
    cin >> n;
    a.assign(n, 0);
    b.assign(n, 0);

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

    dp.assign(n*(1<<n), -1);
    int ans = f(0, 0);
    if(ans == n*n) ans = -1;

    cout << ans << endl;
}