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>

using namespace std;
long long dp[1<<20][20] = {};
int a[20],b[20];
int main(){
	int i,j,k,n;
	long long inf = 10000000000000;
	cin >> n;
	for(i=0;i<n;i++) cin >> a[i];
	for(i=0;i<n;i++) cin >> b[i];
	for(i=0;i<(1<<n);i++){
		for(j=0;j<n;j++){
			dp[i][j] = inf;
		}
	}
	dp[0][0] = 0;
	for(i=0;i<n;i++){
		dp[(1<<i)][i] = 0;
	}
	for(i=0;i<(1<<n);i++){
		for(j=0;j<n;j++){
			if(!((1<<j) & i)) continue;
			int pos = 0;
			for(k=0;k<n;k++){
				if((1<<k) & i) pos++;
			}
			pos--;
			int num,d = (j^pos);
			if(d&1) num = b[j];
			else num = a[j];
			long long cnt = 0;
			for(k=0;k<n;k++){
				if(k==j) continue;
				if(!((1<<k) & i)) continue;
				if(k>j) cnt++;
			}
			for(k=0;k<n;k++){
				if(k==j) continue;
				if(!((1<<k) & i)) continue;
				int num2,d = (pos + 1)^k;
				if(d&1) num2 = b[k];
				else num2 = a[k];
				if(num2<=num){
					dp[i][j] = min(dp[i][j],dp[i^(1<<j)][k] + cnt);
				}
			}
		}
	}
	long long ans = inf;
	for(i=0;i<n;i++){
		ans = min(ans,dp[(1<<n) - 1][i]);
	}
	if(ans==inf){
		cout << -1 << endl;
	}else{
		cout << ans << endl;
	}
}