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# (Mono-mcs 6.8.0.105) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;

namespace contest_Csharp
{
    class Program
    {
        static void Main(string[] args)
        {

            var vals = Console.ReadLine().Split(' ')
                .Select(x => int.Parse(x)).ToList();
            var N = vals[0];
            //var K = vals[1];
            

            var A = Console.ReadLine().Split(' ')
                .Select(x => int.Parse(x)).ToList();
            List<(long, int)> ps = new List<(long, int)>();
            for (int i = 0; i < N; i++)
            {
                ps.Add((A[i], i));
            }
            ps.Sort();
            ps.Reverse();
            List<List<long>> dp=new List<List<long>>();
            dp.Add (new List<long>());
            dp[0].Add(0);
            for (int i = 1; i <= N; i++)
            {
                dp.Add(new List<long>(new long[i + 1]));
                dp[i][0] = dp[i - 1][0] + ps[i-1].Item1 * Math.Abs((N - 1) - (i-1) - ps[i-1].Item2);
                for (int k = 1; k < i; k++)
                {
                    dp[i][k] = Math.Max(dp[i - 1][k - 1] + ps[i-1].Item1 * Math.Abs(ps[i-1].Item2 - (k - 1)),
                       dp[i - 1][k] + ps[i-1].Item1 * Math.Abs((N-1)-(i-1-k)-ps[i-1].Item2));
                }
                dp[i][i] = dp[i - 1][i - 1] + ps[i-1].Item1 * Math.Abs(ps[i-1].Item2 - (i - 1));
            }
            long ans = 0;
            for (int i = 0; i <= N; i++)
            {
                ans = Math.Max(ans, dp[N][i]);
            }
            Console.WriteLine(ans);


        }
       
       

    }
}