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"bits/stdc++.h"
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
#define INF 1e+9


int main() {
	int n,x;
	vector<int> a;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> x;
		a.push_back(x);
	}

	ll ans = 0,ans1=0;
	sort(a.begin(), a.end());
	if (n%2 == 0) { 
		for (int i = 0; i < n / 2; i++) { 
			if (i==(n/2)-1) { ans -= a[i]; }
			else { ans -= (a[i] * 2); }
		}
		for (int i = (n / 2) ; i < n; i++) { 
			if (i == (n / 2) ) { ans += a[i]; }
			else { ans += (a[i] * 2); }
		}
	}
	else {
		for (int i = 0; i < (n / 2) + 1; i++) {
			if (i >=(n/2)-1) { ans -= a[i]; }
			else { ans -= (a[i] * 2); }
		}
		for (int i = (n / 2)+1; i < n; i++) {
			  ans += (a[i] * 2); 
		}
	}

	sort(a.begin(),a.end(),greater<int>());
	int cnt = 0;
	if (n % 2 == 0) {
		
	}
	else {
		for (int i = 0; i < (n / 2) + 1; i++) {
			if (i>=(n/2)-1) { ans1 += a[i];}
			else { ans1 += (a[i] * 2); }
		}
		for (int i = (n / 2) + 1; i < n; i++) {
			ans1 -= (a[i] * 2);
		}
	}
	ans = max(ans, ans1);
	cout << ans << endl;

	return 0;
}