Problem:
Score : 400 points

Problem Statement

You are given N integers; the i-th of them is A_i.
Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.

Constraints

2 \leq N \leq 10^5

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the maximum possible sum of the absolute differences between the adjacent elements after arranging the given integers in a row in any order you like.

Sample Input 1

5
6
8
1
2
3

Sample Output 1

21

When the integers are arranged as 3,8,1,6,2, the sum of the absolute differences between the adjacent elements is |3 - 8| + |8 - 1| + |1 - 6| + |6 - 2| = 21. This is the maximum possible sum.

Sample Input 2

6
3
1
4
1
5
9

Sample Output 2

25

Sample Input 3

3
5
5
1

Sample Output 3

8

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

/**
 * @copyright (c) 2020 Daisuke Hashimoto
 */

#include <bits/stdc++.h>
using namespace std;
using Pair = pair<int64_t, int64_t>;
constexpr int64_t kInf = INT64_MAX / 2L;

// std::cout << std::setprecision(20) << 1.1 << endl;

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

  int64_t N;
  cin >> N;
  // const int64_t L = static_cast<int64_t>(T.length());
  vector<int64_t> A(N);
  for (int64_t i = 0; i < N; ++i) {
    cin >> A[i];
  }
  sort(A.begin(), A.end());

  int64_t result = 0;
  if (N % 2 == 0) {
    int64_t i = 0;
    for (int64_t j = 0; j < (N - 2) / 2; ++j) {
      result -= A[i++] * 2;
    }
    result -= A[i++];
    result += A[i++];
    for (int64_t j = 0; j < (N - 2) / 2; ++j) {
      result += A[i++] * 2;
    }
  } else {
    int64_t result_1 = 0;
    {
      int64_t i = 0;
      for (int64_t j = 0; j < (N - 3) / 2; ++j) {
        result_1 -= A[i++] * 2;
      }
      result_1 -= A[i++];
      result_1 -= A[i++];
      for (int64_t j = 0; j < (N - 1) / 2; ++j) {
        result_1 += A[i++] * 2;
      }
    }

    int64_t result_2 = 0;
    {
      int64_t i = 0;
      for (int64_t j = 0; j < (N - 1) / 2; ++j) {
        result_2 -= A[i++] * 2;
      }
      result_2 += A[i++];
      result_2 += A[i++];
      for (int64_t j = 0; j < (N - 3) / 2; ++j) {
        result_2 += A[i++] * 2;
      }
    }
    result = max(result_1, result_2);
  }

  cout << result << endl;
  return 0;
}