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 <iostream>
#include <vector>

using namespace std;

const int inf = 1<<20;

int dp[1<<18][51];

int main() {
  int n; cin >> n;
  vector<int> la(n), lb(n);
  for (int &a: la) cin >> a;
  for (int &b: lb) cin >> b;
  for (int mask = 0; mask < (1<<n); ++mask) {
    for (int s = 0; s <= 50; ++s) dp[mask][s] = inf;
  }
  dp[0][0] = 0;
  for (int mask = 0; mask < (1<<n); ++mask) {
    int cnt = __builtin_popcount(mask);
    for (int s = 0; s <= 50; ++s) { 
      if (dp[mask][s] == inf) continue;
      int p = 0;
      for (int j = 0; j < n; ++j) { 
        int flag = 1<<j;
        if (mask & flag) continue;
        int t = (abs(j-cnt) % 2) ? lb[j] : la[j];
        if (t >= s) {
          dp[mask|flag][t] = min(dp[mask|flag][t], dp[mask][s] + p);
          // cerr << mask << " " << (mask|flag) << " " << s << " " << t << " " << dp[mask][s] << " " << p << " " << dp[mask|flag][t] << endl;
        }
        ++p;
      }
    }
  }
  int ans = inf;
  for (int t = 1; t <= 50; ++t) {
    ans = min(ans, dp[(1<<n)-1][t]);
  }
  if (ans == inf) {
    cout << -1 << endl;
  }
  else {
    cout << ans << endl;
  }
}