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: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Security.Cryptography;
using System.Text;
using static System.Console;
using static System.Math;

namespace AtTest.ForYellow
{
    class ABC_163
    {
        static void Main(string[] args)
        {
            var sw = new System.IO.StreamWriter(OpenStandardOutput()) { AutoFlush = false };
            SetOut(sw);

            Method(args);

            Out.Flush();
        }

        static void Method(string[] args)
        {
            int n = ReadInt();
            long[] array = ReadLongs();
            long[][] vals = new long[n][];
            for(int i = 0; i < n; i++)
            {
                vals[i] = new long[2] { array[i], i };
            }
            vals = vals.OrderBy(a => -a[0]).ToArray();

            long[,,] dp = new long[n, n+1, n+1];
            dp[0, 1, 0] = vals[0][1] * vals[0][0];
            dp[0, 0, 1] = (n - 1 - vals[0][1]) * vals[0][0];

            for(int i = 1; i < n; i++)
            {
                for(int j = 0; j <= i; j++)
                {
                    long left= dp[i - 1, j, i - j] + Abs(j-vals[i][1]) * vals[i][0];
                    long right = dp[i - 1, j, i - j] + Abs(n - 1 - (i - j)-vals[i][1]) * vals[i][0];
                    dp[i, j + 1, i - j] = Max(dp[i, j + 1, i - j], left);
                    dp[i, j, i - j + 1] = Max(dp[i, j, i - j + 1], right);
                }
            }

            long res = 0;
            for(int i = 0; i <= n; i++)
            {
                res = Max(res, dp[n - 1, i, n - i]);
            }
            WriteLine(res);
        }

        private static string Read() { return ReadLine(); }
        private static char[] ReadChars() { return Array.ConvertAll(Read().Split(), a => a[0]); }
        private static int ReadInt() { return int.Parse(Read()); }
        private static long ReadLong() { return long.Parse(Read()); }
        private static double ReadDouble() { return double.Parse(Read()); }
        private static int[] ReadInts() { return Array.ConvertAll(Read().Split(), int.Parse); }
        private static long[] ReadLongs() { return Array.ConvertAll(Read().Split(), long.Parse); }
        private static double[] ReadDoubles() { return Array.ConvertAll(Read().Split(), double.Parse); }
    }
}