You are given a chessboard of size 1 × n. It is guaranteed that n is even. The chessboard is painted like this: "BWBW...BW".

Some cells of the board are occupied by the chess pieces. Each cell contains no more than one chess piece. It is known that the total number of pieces equals to <image>.

In one step you can move one of the pieces one cell to the left or to the right. You cannot move pieces beyond the borders of the board. You also cannot move pieces to the cells that are already occupied.

Your task is to place all the pieces in the cells of the same color using the minimum number of moves (all the pieces must occupy only the black cells or only the white cells after all the moves are made).

Input

The first line of the input contains one integer n (2 ≤ n ≤ 100, n is even) — the size of the chessboard. 

The second line of the input contains <image> integer numbers <image> (1 ≤ pi ≤ n) — initial positions of the pieces. It is guaranteed that all the positions are distinct.

Output

Print one integer — the minimum number of moves you have to make to place all the pieces in the cells of the same color.

Examples

Input

6
1 2 6


Output

2


Input

10
1 2 3 4 5


Output

10

Note

In the first example the only possible strategy is to move the piece at the position 6 to the position 5 and move the piece at the position 2 to the position 3. Notice that if you decide to place the pieces in the white cells the minimum number of moves will be 3.

In the second example the possible strategy is to move <image> in 4 moves, then <image> in 3 moves, <image> in 2 moves and <image> in 1 move.
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
long long power(long long a, long long b) {
  long long result = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      result *= a;
    }
    a *= a;
    b /= 2;
  }
  return result;
}
long long gcd(long long x, long long y) {
  long long r;
  while (y != 0 && (r = x % y) != 0) {
    x = y;
    y = r;
  }
  return y == 0 ? x : y;
}
long long countSetBits(long long x) {
  long long Count = 0;
  while (x > 0) {
    if (x & 1) Count++;
    x = x >> 1;
  }
  return Count;
}
bool isPerfectSquare(long long n) {
  long long sr = sqrt(n);
  if (sr * sr == n)
    return true;
  else
    return false;
}
long long mod(long long x, long long M) { return ((x % M + M) % M); }
long long add(long long a, long long b, long long M) {
  return mod(mod(a, M) + mod(b, M), M);
}
long long mul(long long a, long long b, long long M) {
  return mod(mod(a, M) * mod(b, M), M);
}
long long powerM(long long a, long long b, long long M) {
  long long res = 1ll;
  while (b) {
    if (b % 2ll == 1ll) {
      res = mul(a, res, M);
    }
    a = mul(a, a, M);
    b /= 2ll;
  }
  return res;
}
long long mod_inv(long long a, long long m) {
  long long g = m, r = a, x = 0, y = 1;
  while (r != 0) {
    long long q = g / r;
    g %= r;
    swap(g, r);
    x -= q * y;
    swap(x, y);
  }
  return mod(x, m);
}
long long nCr(long long n, long long k) {
  if (n < k) return 0;
  if (k == 0) return 1;
  long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long modInverse(long long n, long long M) { return powerM(n, M - 2, M); }
vector<long long> factu;
void factcalc(long long n, long long M) {
  factu.resize(n + 1);
  factu[0] = 1;
  for (long long i = 1; i <= n; i++) {
    factu[i] = mul(factu[i - 1], i, M);
  }
}
long long nCrM(long long n, long long r, long long M) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return mul(mul(factu[n], modInverse(factu[r], M), M),
             modInverse(factu[n - r], M), M);
}
long long rand_int(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(rng);
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n / 2);
  for (long long i = 0; i < n / 2; i++) {
    cin >> v[i];
  }
  sort(v.begin(), v.end());
  long long a1 = 0, a2 = 0;
  for (long long i = 1; i <= n; i += 2) {
    a1 += abs(i - v[i / 2]);
  }
  for (long long i = 2; i <= n; i += 2) {
    a2 += abs(i - v[(i - 1) / 2]);
  }
  cout << min(a1, a2) << endl;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(20);
  solve();
  return 0;
}
