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 C# (.NET Core 3.1.201) (judge verdict: Accepted)

using System.Linq;
using static System.Math;
using System;

public class P
{
    public long act { get; set; }
    public int id { get; set; }
}

public class Hello
{
    public static void Main()
    {
        var n = int.Parse(Console.ReadLine().Trim());
        string[] line = Console.ReadLine().Trim().Split(' ');
        var a = Array.ConvertAll(line, long.Parse);
        var ps = new P[n];
        for (int i = 0; i < n; i++)
            ps[i] = new P { id = i, act = a[i] };
        ps = ps.OrderByDescending(x => x.act).ToArray();
        goDP(n, ps);
    }
    static void goDP(int n, P[] ps)
    {
        var dp = new long[n + 1, n + 1];
        for (int i = 0; i < n; i++)
            for (int j = 0; j <= i; j++)
            {
                var te = dp[i, j] + calcL(j, ps[i]);
                dp[i + 1, j + 1] = Max(dp[i + 1, j + 1], te);
                var te2 = dp[i, j] + calcR(n, i, j, ps[i]);
                dp[i + 1, j] = Max(dp[i + 1, j], te2);
            }
        var ans = 0L;
        for (int i = 1; i < n + 1; i++)
            ans = Max(ans, dp[n, i]);
        Console.WriteLine(ans);
    }
    static long calcL(int y, P p) => p.act * Abs(y - p.id);
    static long calcR(int n, int x, int y, P p) => p.act * Abs(n - (x - y) - 1 - p.id);
}