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: Wrong Answer)

#include <iostream>
#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <regex>
#include <bitset>
#include <string>
#include <vector>
#include <iomanip>
#include <sstream>
#include <utility>
#include <iterator>
#include <algorithm>
#include <functional>
#define rep(i, n) for(int i = 0;i < n;i++)
#define REP(i, n) for(int i = 2;i < n;i++)
#define ll long long

using namespace std;

ll n, m, o, add = 1, A[100001], B[100001], ans[100001];
deque<ll> d;
int main() {
	cin >> n;
	rep(i, n) {
		cin >> A[i];
	}
	sort(A, A + n);
	rep(i, n) {
		if (i % 2 == 0) B[i] = A[i / 2];
		else B[i] = A[n - i / 2 - 1];
	}
	d.push_front(B[0]);
	d.push_back(B[1]);
	if (n % 2 == 0) {
		n--;
		m = B[n];
		n--;
		o = B[n];
	}
	else {
		n--;
	}
	REP(i, n) {
		if (i % 2 == 0) {
			d.push_back(B[i]);
		}
		else {
			d.push_front(B[i]);
		}
	}
	if (o != 0) {
		if (d.begin() > d.end()) {
			d.push_back(o);
			d.push_front(m);
		}
		else {
			d.push_back(m);
			d.push_front(o);
		}
	}
	else {
		d.begin() > d.end() ? d.push_back(m) : d.push_front(m);
	}
	long long sum = 0;
	rep(i, n) {
		if (i != 0) {
			sum += abs(d[i] - d[i - 1]);
		}
	}
	cout << sum << endl;
	return 0;
}