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: Time Limit Exceeded)


n = parse(Int, readline())
arr = [tpl for tpl in enumerate(parse.(Int, split(readline())))]
arr = sort(arr, by = tpl -> -tpl[2])

dp = Dict()

for ln = 0 : n
  for rn = 0 : n - ln
    if ln + rn > n || get(dp, (ln, rn), 0) > 0
      continue
    end

    if ln + rn > 0
      p, a = arr[ln + rn]
      val_l = ln > 0 ? dp[(ln - 1, rn)] + a * abs(p - ln) : 0
      val_r = rn > 0 ? dp[(ln, rn - 1)] + a * abs(p - (n - rn + 1)) : 0

      dp[(ln, rn)] = max(val_l, val_r)
    else
      dp[(0, 0)] = 0
    end
  end
end

println(maximum([dp[i, n - i] for i in 0 : n]))