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 Ruby (2.7.1) (judge verdict: Runtime Error)

N = gets.to_i
A = gets.chomp.split.map(&:to_i)
left = 0
right = N-1

B = A.map.with_index { |elem, index| [index, elem]}
C = B.sort_by { |b| b[1] }.reverse
dpn = { [0,0] => 0 }
dpn1 = {}
(1..N).each do |n|
  dpn.each do |d|
    
    # left
    n_left = left + d[0][0]
    if C[n-1][0] >= n_left and (d[0][0]+1-d[0][1]).abs <= (N-n)
      if dpn1[[d[0][0]+1,d[0][1]]].nil?
        dpn1[[d[0][0]+1,d[0][1]]] = d[1] + C[n-1][1] * (C[n-1][0] - n_left)
      else
        dpn1[[d[0][0]+1,d[0][1]]] = [d[1] + C[n-1][1] * (C[n-1][0] - n_left), dpn1[[d[0][0]+1,d[0][1]]]].max
      end
    end
    
    # right
    n_right = right - d[0][1]
    if C[n-1][0] <= n_right and (d[0][0]-(d[0][1]+1)).abs <= (N-n)
      if  dpn1[[d[0][0],d[0][1]+1]].nil?
        dpn1[[d[0][0],d[0][1]+1]] = r = d[1] + C[n-1][1] * (C[n-1][0] - n_right).abs
      else
        dpn1[[d[0][0],d[0][1]+1]] = [r = d[1] + C[n-1][1] * (C[n-1][0] - n_right).abs, dpn1[[d[0][0],d[0][1]+1]]].max
      end
    end
  end
  dpn = dpn1
  # p dpn
  dpn1 = {}
end
p dpn.max { |x, y| x[1] <=> y[1] }[1]