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;

int N;
int A[18][2];
int dp[1 << 18][18];

int main() {
   ios_base::sync_with_stdio(false);
   cin >> N;
   for (int i = 0; i < N; ++i) cin >> A[i][0];
   for (int i = 0; i < N; ++i) cin >> A[i][1];
   for (int i = 1; i < N; i += 2) swap(A[i][0], A[i][1]);
   for (int mask = 0; mask < (1 << N); ++mask) {
      for (int last = 0; last < N; ++last) {
         dp[mask][last] = 1 << 30;
      }
   }
   for (int i = 0; i < N; ++i) {
      dp[1 << i][i] = 0;
   }
   for (int mask = 1; mask + 1 < (1 << N); ++mask) {
      int cnt = __builtin_popcount(mask);
      for (int last = 0; last < N; ++last) {
         if (dp[mask][last] == (1 << 30)) continue;
         for (int nxt = 0; nxt < N; ++nxt) if (!(mask & (1 << nxt)) && A[last][cnt & 1 ^ 1] <= A[nxt][cnt & 1]) {
            dp[mask | (1 << nxt)][nxt] = min(dp[mask | (1 << nxt)][nxt], dp[mask][last] + cnt - __builtin_popcount(mask & ((1 << nxt) - 1)));
         }
      }
   }
   int ans = *min_element(dp[(1 << N) - 1], dp[(1 << N) - 1] + N);
   if (ans == (1 << 30)) {
      ans = -1;
   }
   cout << ans << "\n";
   return 0;
}