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 <stdio.h>
typedef long long ll;

int main(void) {
  ll i, j, k, n, m, cnt, now, now2, ans = 1e9, sum;
  scanf("%lld", &n);
  m = 1 << n;
  ll a[n], b[n], dp[m][n];
  for(i = 0; i < n; ++i) scanf("%lld", &a[i]);
  for(i = 0; i < n; ++i) scanf("%lld", &b[i]);
  for(i = 0; i < m; ++i) for(j = 0; j < n; ++j) dp[i][j] = 1e9;
  for(i = 0; i < n; ++i) dp[1 << i][i] = i;
  for(i = 1; i < m; ++i) {
    cnt = 0;
    for(j = 0; j < n; ++j) if((i >> j) & 1) ++cnt;
    for(j = 0; j < n; ++j) {
      now = ((cnt & 1) == (j & 1) ? b[j] : a[j]);
      sum = 0;
      for(k = 0; k < n; ++k) if(!((i >> k) & 1)) {
        now2 = ((cnt & 1) == (k & 1) ? a[k] : b[k]);
        if(now <= now2 && dp[i][j] + sum < dp[i | (1 << k)][k])
          dp[i | (1 << k)][k] = dp[i][j] + sum;
        sum++;
      }
    }
  }
  for(i = 0; i < n; ++i) if(ans > dp[m - 1][i]) ans = dp[m - 1][i];
  printf("%lld", ans == 1e9 ? -1 : ans);
  return 0;
}