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>
#define rep(i,n) for(int i=(0);i<(n);i++)

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	int n;
	cin >> n;

	int a[n], b[n];
	rep(i, n) cin >> a[i];
	rep(i, n) cin >> b[i];

	int max_bit = (1 << n) - 1;
	int max_a = 52;
	int dp[max_bit + 1][max_a + 1];

	int INF = 1010101;
	rep(i, max_bit + 1) rep(j, max_a+1) dp[i][j] = INF;
	dp[0][0] = 0;

	rep(bit, max_bit + 1){
		int cnt = __builtin_popcount(bit);

		int t = cnt;
		vector<pii> v;

		rep(i, n){
			if(!((bit >> i) & 1)){
				v.push_back({i, t++});
			}
		}
		
		rep(j, max_a + 1){
			if(dp[bit][j] == INF) continue;

			for(pii p : v){
				int ns;
				if(abs(p.first - cnt) % 2 == 0){
					ns = a[p.first];
				}else{
					ns = b[p.first];
				}

				if(ns >= j){
					dp[bit | (1 << p.first)][ns] 
				     = min(dp[bit | (1 << p.first)][ns], dp[bit][j] + (p.second - cnt));
				}
				
			}
			
		}
	}

	int ans = INF;
	rep(i, max_a+1) ans = min(ans, dp[max_bit][i]);
	if(ans == INF) ans = -1;

	cout << ans << endl;
}