Problem:
Score : 500 points

Problem Statement

There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.

You can rearrange these children just one time in any order you like.

When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.

Find the maximum total happiness points the children can earn.

Constraints

2 \leq N \leq 2000

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_2 ... A_N

Output

Print the maximum total happiness points the children can earn.

Sample Input 1

4
1 3 4 2

Sample Output 1

20

If we move the 1-st child from the left to the 3-rd position from the left, the 2-nd child to the 4-th position, the 3-rd child to the 1-st position, and the 4-th child to the 2-nd position, the children earns 1 \times |1-3|+3 \times |2-4|+4 \times |3-1|+2 \times |4-2|=20 happiness points in total.

Sample Input 2

6
5 5 6 1 1 1

Sample Output 2

58

Sample Input 3

6
8 6 9 1 2 1

Sample Output 3

85

Solution in Julia (1.4.0) (judge verdict: Wrong Answer)

parseInt(x)=parse(Int, x)
parseMap(x::Array{SubString{String},1})=map(parseInt, x)

function main()
	n = readline() |> parseInt
	a = readline() |> split |> parseMap
	b = Tuple{Int,Int,Int}[]
	for i in 1:n
		for j in 1:n
			push!(b,(i,j,a[i]*abs(i-j)))
		end
	end
	b = sort(b,by=x->x[3],rev=true)
	ans = -ones(Int,n)
	used = zeros(Int,n)
	for i in 1:n*n
		(p,q,r) = b[i]
		if ans[q]==-1&&used[p]==0
			used[p]=1
			ans[q]=r
		end
	end
	println(sum(ans))
end

main()