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 Lua (Lua 5.3.5) (judge verdict: Accepted)

local mmi, mma = math.min, math.max
local mab = math.abs
local n = io.read("*n")
local t = {}
for i = 1, n do
  t[i] = {io.read("*n")}
  t[i][2] = i
end
table.sort(t, function(a, b) return a[1] > b[1] end)
local dp = {}
for i = 1, n do
  local val, idx = t[i][1], t[i][2]
  dp[i] = {}
  for j = 1, i + 1 do
    -- j: left use count - 1
    dp[i][j] = 0
  end
  if i == 1 then
    dp[i][1] = val * (n - idx)
    dp[i][2] = val * (idx - 1)
  else
    for j = 1, i do
      -- dst: j
      dp[i][j + 1] = mma(dp[i][j + 1], dp[i - 1][j] + val * mab(idx - j))
      -- dst: n - (i - (j - 1)) + 1
      dp[i][j] = mma(dp[i][j], dp[i - 1][j] + val * mab(n - i + j - idx))
    end
  end
  -- print(table.concat(dp[i], " "))
end
local ret = 0
for i = 1, n + 1 do
  ret = mma(ret, dp[n][i])
end
print(ret)