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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;

public class Main {

	private static boolean isOdd(int val) {
		return ((val & 1) != 0);
	}

	public static void main(String[] args) {
		try (Scanner in = new Scanner(System.in)) {
			final int N = in.nextInt();
			final long[] A = new long[N];
			for (int i = 0; i < N; ++i) {
				A[i] = in.nextInt();
			}

			final int halfN = N / 2;
			long result = 0L;
			Arrays.sort(A);
			if (isOdd(N)) {

				long r1 = 0;
				r1 -= 2 * Arrays.stream(A, 0, halfN).sum();
				r1 += 2 * Arrays.stream(A, halfN, N).sum();
				r1 -= A[halfN];
				r1 -= A[halfN + 1];

				long r2 = 0;
				r2 -= 2 * Arrays.stream(A, 0, halfN + 1).sum();
				r2 += 2 * Arrays.stream(A, halfN + 1, N).sum();
				r2 += A[halfN];
				r2 += A[halfN - 1];

				result = Math.max(r1, r2);

			} else {
				result -= 2 * Arrays.stream(A, 0, halfN).sum();
				result += 2 * Arrays.stream(A, halfN, N).sum();
				result -= A[halfN];
				result += A[halfN - 1];
			}
			System.out.println(result);
		}
	}
}