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: Accepted)

n = gets.to_i
a = gets.split.map(&:to_i)

ary = []
a.each_with_index{|x, i| ary.push([x, i]) }

ary = ary.sort{|lhs, rhs| rhs[0] <=> lhs[0] }

left = 0
right = 0

dp = Array.new(n) { Array.new(n + 1) { 0 } }

ary.each_with_index do |el, i|
  (0..i).each do |j|
    left = j
    right = i - j
    
    current = 0
    current = dp[i - 1][j] if i > 0
    
    if el[1] - left >= 0
      dp[i][left + 1] = [current + (el[1] - left) * el[0], dp[i][left + 1]].max
    end
    if (n - right - 1) - el[1] >= 0
      dp[i][left] = [current + ((n - right - 1) - el[1]) * el[0], dp[i][left]].max
    end
  end
end

puts dp[n - 1].max