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;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (n); i++)

void chmin(int &x, int y) {
  x = min(x, y);
}

int popcount(int x) {
  int res = 0;
  while (x > 0) {
    res++;
    x &= x - 1;
  }
  return res;
}

int main() {
  int N; cin >> N;
  vector<int> A(N); rep(i, N) cin >> A[i];
  vector<int> B(N); rep(i, N) cin >> B[i];
  vector<vector<int>> dp(1 << N, vector<int>(N, INT_MAX));
  dp[0][0] = 0;
  auto f = [&](int i, int j) {
    return abs(j - i) % 2 == 0 ? A[i] : B[i];
  };
  rep(i, 1 << N) rep(j, N) if (dp[i][j] != INT_MAX) {
    int pos = popcount(i);
    rep(k, N) if (~i >> k & 1) {
      int cost = 0;
      rep(l, N) if ((i >> l & 1) and k < l) cost++;
      if (i == 0 or f(j, pos - 1) <= f(k, pos)) {
        chmin(dp[i | 1 << k][k], dp[i][j] + cost);
      }
    }
  }
  int ans = INT_MAX;
  rep(j, N) chmin(ans, dp[(1 << N) - 1][j]);
  if (ans == INT_MAX) ans = -1;
  cout << ans << endl;
}