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;
using P = pair<int, int>;
#define all(c) (c).begin(), (c).end()
const int INF = 1e9 + 1;
const ll LINF = 1e18 + 1;
const ll MOD = 1e9 + 7;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int N;
  cin >> N;
  vector<int> a(N), b(N);
  vector<vector<int>> dp(1 << N, vector<int>(51, INF));
  for (int i = 0; i < N; ++i) cin >> a[i];
  for (int i = 0; i < N; ++i) cin >> b[i];

  dp[0][0] = 0;

  for (int i = 0; i < (1 << N); ++i) {
    vector<int> cnt(N, 0);
    for (int k = N - 2; k >= 0; --k) {
      cnt[k] = cnt[k + 1];
      if ((i >> (k + 1)) % 2 == 1) ++cnt[k];
    }

    int usedCnt = 0, tmp = i;
    while (tmp > 0) {
      if (tmp % 2) ++usedCnt;
      tmp /= 2;
    }
    for (int j = 0; j < N; ++j) {
      if ((i >> j) % 2 == 1) continue;
      int nxtBit = i + (1 << j);

      int nxtNum = (usedCnt % 2 == j % 2) ? a[j] : b[j];
      for (int k = 0; k <= nxtNum; ++k) {
        dp[nxtBit][nxtNum] = min(dp[i][k] + cnt[j], dp[nxtBit][nxtNum]);
      }
    }
  }

  int ans = INF;
  for (int i = 0; i < 51; ++i) {
    ans = min(ans, dp[(1 << N) - 1][i]);
  }
  if (ans == INF) ans = -1;
  cout << ans << endl;
  return 0;
}