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>

#ifdef NON_SUBMIT
#define TEST(n) (n)
#define tout cerr
#else
#define TEST(n) ((void)0)
#define tout cin
#endif

using namespace std;

int D[51][1<<18], A[18], B[18];
bool chk[51][1<<18];

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
	TEST(freopen("input.txt","r",stdin));
	TEST(freopen("output.txt","w",stdout));
	TEST(freopen("debug.txt","w",stderr));
	int N, ans=0x7fffffff;
	queue<pair<int,int>> Q;
	memset(D,0x7f,sizeof(D));
	cin>>N;
	for(int i=0;i<N;i++) cin>>A[i];
	for(int i=0;i<N;i++) cin>>B[i];
	D[0][0]=0;
	Q.emplace(0,0);
	while(!Q.empty()) {
		int v, b, cnt, nv, ncnt=0;
		tie(v,b)=Q.front();
		Q.pop();
		if(b==(1<<N)-1) ans=min(ans,D[v][b]);
		cnt=__builtin_popcount(b);
		for(int i=0;i<N;i++) if(~b&(1<<i)) {
			nv=((i&1)^(cnt&1)) ? B[i]:A[i];
			if(v<=nv && D[nv][b|(1<<i)]>D[v][b]+ncnt) {
				D[nv][b|(1<<i)]=D[v][b]+ncnt;
				if(!chk[nv][b|(1<<i)]) {
					chk[nv][b|(1<<i)]=true;
					Q.emplace(nv,b|(1<<i));
				}
			}
			ncnt++;
		}
	}
	cout<<(ans==0x7fffffff ? -1:ans)<<'\n';
	return 0;
}