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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
  ll n;cin>>n;
  vector<ll> a(n);
  for(ll i=0;i<n;i++)cin>>a[i];
  sort(a.begin(),a.end());
  ll m=0;
  if(n==2){cout<<abs(a[0]-a[1])<<endl;
return 0;
}
 
  for(ll j=0;j<3;j++){
    vector<ll>b(n);
    for(ll k=0;k<n;k++)b[k]=a[k];
    vector<ll>ans={};
    ans.push_back((n+1)/2-1+j);
  b.erase(b.begin()+(n+1)/2-1+j);
  for(ll i=0;i<n-1;i++){
    ll t=b[0];
    ll r=b[b.size()-1];
    ll e=ans[ans.size()-1];
    if(abs(t-e)>abs(r-e)){
      ans.push_back(t);
      b.erase(b.begin());
    }
    else{
      ans.push_back(r);
      b.erase(b.begin()+b.size()-1);
    }
  }
   ll sum=0;
  for(ll i=0;i<ans.size()-1;i++)sum+=abs(ans[i]-ans[i+1]);
  m=max(m,sum);
  }
  cout<<m<<endl;
}