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;

const int N = 18, K = 51, inf = 1e7 + 7;

int dp[1 << N][K];

int a[N], b[N];

int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i)
    scanf("%d", a+i);
  for (int i = 0; i < n; ++i)
    scanf("%d", b+i);
  int nmask = 1 << n;
  for (int i = 0; i < nmask; ++i) {
    for (int j = 0; j < K; ++j)
      dp[i][j] = inf;
  }
  dp[0][0] = 0;
  for (int mask = 0; mask < nmask; ++mask) {
    int cost = 0, odd = __builtin_popcount(mask) & 1;
    for (int i = 0; i < n; ++i) {
      if (mask & (1 << i)) continue;
      int val = ((odd^i) & 1) ? b[i] : a[i];
      int new_mask = mask | (1 << i);
      for (int j = 0; j <= val; ++j) {
        dp[new_mask][val] = min(dp[new_mask][val], cost + dp[mask][j]);
      }
      ++cost;
    }
  }
  int ans = inf;
  for (int i = 0; i < K; ++i)
    ans = min(ans, dp[nmask-1][i]);
  if (ans >= inf)
    ans = -1;
  printf("%d\n", ans);
  return 0;
}