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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main(){
  int N;
  cin >> N;
  priority_queue<int64_t> mapq;
  priority_queue<int64_t, vector<int64_t>, greater<int64_t> > mipq;
  int64_t A;
  for(int i=0;i<N;i++){
    cin >> A;
    mapq.push(A);
    mipq.push(A);
  }
  if(N==2){
    cout << mapq.top()-mipq.top() << endl;
    return 0;
  }
  int count=0;
  int64_t ans=0;
  vector<int64_t> ma(2);
  vector<int64_t> mi(2);
  ma.at(0)=mapq.top();
  mapq.pop();
  count++;
  ma.at(1)=mapq.top();
  mapq.pop();
  count++;
  ans+=ma.at(0)+ma.at(1)-mipq.top()*2;
  mipq.pop();
  count++;
  while(count<N){
    bool flag=false;
    for(int i=0;i<2;i++){
      if(i==1){
        if(mipq.top()==ma.at(i)){
          continue;
        }
      }
      mi.at(i)=mipq.top();
      mipq.pop();
      count++;
      ans+=ma.at(i)-mi.at(i);
      if(count>=N){
        flag=true;
        break;
      }
    }
    if(flag){
      break;
    }
    for(int i=0;i<2;i++){
      if(i==1){
        if(mapq.top()==mi.at(i)){
          continue;
        }
      }
      ma.at(i)=mapq.top();
      mapq.pop();
      count++;
      ans+=ma.at(i)-mi.at(i);
      if(count>=N){
        break;
      }
    }
  }
  cout << ans << endl;
  return 0;
}