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<cstdio>
#include<algorithm>
using namespace std;
int gi(){
	int x=0,w=1;char ch=getchar();
	while ((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if (ch=='-') w=0,ch=getchar();
	while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return w?x:-x;
}
#define ll long long
const int N = 1e5+5;
int n,a[N];
ll cal(int m,int op1,int op2){
	ll sum=0;
	for (int i=1;i<=m;++i) sum-=a[i]<<1;
	sum+=a[m+1]*op1;sum+=a[m+2]*op2;
	for (int i=m+3;i<=n;++i) sum+=a[i]<<1;
	return sum;
}
int main(){
	n=gi();
	for (int i=1;i<=n;++i) a[i]=gi();
	sort(a+1,a+n+1);
	if (n&1) printf("%lld\n",max(cal(n-3>>1,-1,-1),cal(n-1>>1,1,1)));
	else printf("%lld\n",cal(n-2>>1,-1,1));
	return 0;
}