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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <iomanip>
#include <string>
#include <cmath>
#include <algorithm>
#include <cassert>
#include <vector>
#include <array>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <numeric>
using ll = long long;
using namespace std;
using pii = pair<int, int>;

int N;
vector<int> A;

int main() {
	cin >> N;
	for (int i = 0; i < N; i++)
	{
		int tmpA;
		cin >> tmpA;
		A.push_back(tmpA);
	}
	sort(A.begin(), A.end());

	ll sum = 0;
	if (N % 2 == 0) {
		for (int i = 0; i < N / 2 - 1; i++)
		{
			sum -= A.at(i) * 2;
		}
		sum -= A.at(N / 2 - 1);
		sum += A.at(N / 2);
		for (int i = N / 2 + 1; i < N; i++)
		{
			sum += A.at(i) * 2;
		}
	}
	else {
		for (int i = 0; i < (N - 1) / 2 - 1; i++)
		{
			sum -= A.at(i) * 2;
		}
		sum -= A.at((N - 1) / 2 - 1);
		sum -= A.at((N - 1) / 2);
		for (int i = (N - 1) / 2 + 1; i < N; i++)
		{
			sum += A.at(i) * 2;
		}

		ll sum2 = 0;
		for (int i = 0; i < (N - 1) / 2; i++)
		{
			sum2 -= A.at(i) * 2;
		}
		sum2 += A.at((N - 1) / 2);
		sum2 += A.at((N - 1) / 2 + 1);
		for (int i = (N - 1) / 2 + 2; i < N; i++)
		{
			sum2 += A.at(i) * 2;
		}

		sum = max(sum, sum2);
	}


	cout << sum << endl;
	return 0;
}