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 Fortran (gfortran v4.8.4) (judge verdict: Runtime Error)

program test
implicit none

integer(8) :: n,a(200000),dummy(200000)=0,i,j,ans,k,ans1,ans2,ans3

read(*,*) n
do i=1,n
	read(*,*) a(i)
enddo

call heapsort_down(n,a(1:n),dummy(1:n))

if(n==2)then
	ans = abs(a(2)-a(1))
else if(n==3) then
	ans1 = abs(a(1)-a(2))+abs(a(2)-a(3))
	ans2 = abs(a(1)-a(3))+abs(a(3)-a(2))
	ans3 = abs(a(3)-a(1))+abs(a(1)-a(2))
	ans = max(ans1,ans2,ans3)
else if(mod(n,2)==0) then !偶数
	k = n/2
	ans = 2*sum(a(1:k-1)) + a(k) - a(k+1) -2*sum(a(k+2:n))
else !奇数
	k = (n-1)/2
	ans = 2*sum(a(1:k)) - a(k+1) - a(k+2) -2*sum(a(k+3:n))
	ans = ans / 0
endif

write(*,*) ans

contains
subroutine heapsort_down(n,y1,y2)
 
	implicit none
	integer(8),intent(in) :: n
	integer(8),intent(inout) :: y1(1:n)
	integer(8),intent(inout) :: y2(1:n)
	integer(8) :: i,k,j,l
	integer(8) :: t1,t2
	
	l = n/2+1
	k = n
 
	do while (k /= 1)
		 if(l > 1) then
				l = l-1
				t1 = y1(l)
				t2 = y2(l)
				
		 else
				t1 = y1(k)
				t2 = y2(k)
				y1(k) = y1(1)
				y2(k) = y2(1)
				k = k-1
				if(k == 1) then
					 y1(1) = t1
					 y2(1) = t2
					 exit
				endif
		 endif
 
		 i = l
		 j = l+l
 
		 do while(j <= k)
				if(j < k) then
					 if(y1(j) > y1(j+1)) j = j+1
				endif
				
				if (t1 > y1(j)) then
					 y1(i) = y1(j)
					 y2(i) = y2(j)
					 i = j
					 j = j+j
				else
					 j = k+1
				endif
		 
		 enddo
		 
		 y1(i) = t1
		 y2(i) = t2
	
	enddo
	return
 
end subroutine heapsort_down
 
subroutine heapsort_up(n,y1,y2)
 
	implicit none
	integer(8),intent(in) :: n
	integer(8),intent(inout) :: y1(1:n)
	integer(8),intent(inout) :: y2(1:n)
	integer(8) :: i,k,j,l
	integer(8) :: t1,t2
	
	l = n/2+1
	k = n
 
	do while (k /= 1)
		 if(l > 1) then
				l = l-1
				t1 = y1(l)
				t2 = y2(l)
				
		 else
				t1 = y1(k)
				t2 = y2(k)
				y1(k) = y1(1)
				y2(k) = y2(1)
				k = k-1
				if(k == 1) then
					 y1(1) = t1
					 y2(1) = t2
					 exit
				endif
		 endif
 
		 i = l
		 j = l+l
 
		 do while(j <= k)
				if(j < k) then
					 if(y1(j) < y1(j+1)) j = j+1
				endif
				
				if (t1 < y1(j)) then
					 y1(i) = y1(j)
					 y2(i) = y2(j)
					 i = j
					 j = j+j
				else
					 j = k+1
				endif
		 
		 enddo
		 
		 y1(i) = t1
		 y2(i) = t2
	
	enddo
	return
 
end subroutine heapsort_up

end program