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;

typedef long long ll;

double getTime() { 
  return clock() / (double) CLOCKS_PER_SEC;
}

void read() {};

template<typename T, typename... Args>
void read(T& a, Args&... args) {
  cin >> a;
  read(args...);
}

void print() {};

template <typename T, typename... Args>
void print(T a, Args... args) {
  cout << a << " \n"[sizeof...(args) == 0];
  print(args...);
}

int dp[1 << 18][51];
int n;
vector <int> a, b;

int go(int mask, int last) {
  if(mask == (1 << n) - 1) return 0;
  if(dp[mask][last] != -1) return dp[mask][last];
  int ans = 1e9 + 10;
  int idx = 0; int cur = __builtin_popcount(mask);
  for(int i = 0; i < n; i++) {
    if((mask >> i) & 1) {
      cur++;
      continue;
    }
    if((cur + idx) & 1) {
      if(b[i] >= last) ans = min(ans, idx + go(mask | (1 << i), b[i]));
    }
    else {
      if(a[i] >= last) ans = min(ans, idx + go(mask | (1 << i), a[i]));
    }
    idx++;
  }
  return dp[mask][last] = ans;
}

int main() {
  read(n);
  a.resize(n), b.resize(n);
  for(int i = 0; i < n; i++) {
    read(a[i]);
  }
  for(int i = 0; i < n; i++) {
    read(b[i]);
  }
  memset(dp, -1, sizeof dp);
  int ans = go(0, 0);
  if(ans >= 1000000000) ans = -1;
  print(ans);
}