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;
inline int read(){
    int x = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0',c = getchar();
    return x;
}

const int N = 18 + 5;
int n,a[N],b[N],ans;
int f[19][1<<18],g[19][1<<18];
int L;
inline void upd(int &x,int v){ v < x ? x = v : 0; }
int val[N][N];
inline void DP(int x){
	int i,j,k,cnt;
	for (i = 0; i <= n; ++i)
	for (j = 0; j < L; ++j) g[i][j] = f[i][j],f[i][j] = 1000000;
	for (i = 0; i <= n; ++i)
	for (j = 0; j < L; ++j) if (g[i][j]<=10000){
		cnt = 0;
		for (k = 1; k <= n; ++k) if (!(j>>(k-1)&1)){
			if (val[i][x-1] <= val[k][x]) upd(f[k][j|(1<<k-1)],g[i][j]+cnt);
			++cnt;
		}
	}
}
int main(){
	int i,j;
	ans = 1000000;
	n = read(),L = 1<<n;
	for (i = 1; i <= n; ++i) a[i] = read();
	for (i = 1; i <= n; ++i) b[i] = read();
	for (i = 1; i <= n; ++i)
	for (j = 1; j <= n; ++j) val[i][j] = ((i^j)&1) ? (b[i]) : (a[i]);
	for (i = 0; i <= n; ++i) for (j = 0; j < L; ++j) f[i][j] = 1000000;
	f[0][0] = 0;
	for (i = 1; i <= n; ++i) DP(i);
	for (i = 1; i <= n; ++i) ans = min(ans,f[i][(1<<n)-1]);
	if (ans > 10000) ans = -1;
	cout << ans << endl;
    return 0;
}