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 <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define print(s) cout << s << endl
#define acc(v) accumulate(v.begin(), v.end(), 0)
#define cinv(n,v) rep(i,n) cin >> v[i]
using namespace std;
const int INF = 1e9;
typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vint;
typedef vector<string> vstr;
typedef vector<char> vchar;
const ll LINF = 1e18;
const ll MOD = 1e9 + 7;
// const lint MOD = 998244353;
int x_pos[4] = {1,0,-1,0}, y_pos[4] = {0,1,0,-1};
int ctoi(char c) {
	if (c >= '0' && c <= '9') {
		return c - '0';
	}
	return 0;
}

char upper(char c){
	return c-0x20;
}

char lower(char c){
	return c+0x20;
}

void unique_vector(vector<int>& v){
	sort(all(v));
	v.erase(unique(all(v)), v.end());
}

// n次元配列の初期化。第２引数の型のサイズごとに初期化していく。
template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val){
    std::fill( (T*)array, (T*)(array+N), val );
}

int main() {
    int n;cin>>n;
    vector<int> v(n);
    rep(i,n)cin>>v[i];
    sort(all(v));
    vector<int> ans(n);
    ans[n/2] = v[0];
    rep(i,(n-1)/2){
        if(i%2==0){
            ans[n/2-i-1] = v[n-1-i/2];
            ans[n/2+i+1] = v[n-1-i/2-1];
        }else{
            ans[n/2+i+1] = v[i+1];
            ans[n/2-i-1] = v[i];
        }
    }
    if(n%2==0)ans[0] = v[n/2];
    ll Ans =  0;
    rep(i,n-1){
        Ans += abs(ans[i]-ans[i+1]);
    }
    cout << Ans << endl;
}