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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        EActiveInfants solver = new EActiveInfants();
        solver.solve(1, in, out);
        out.close();
    }

    static class EActiveInfants {
        pair[] arr;
        int n;
        long[][] memo;

        public void solve(int testNumber, Scanner sc, PrintWriter pw) {
            n = sc.nextInt();
            arr = new pair[n];
            for (int i = 0; i < n; i++)
                arr[i] = new pair(sc.nextInt(), i);
            Arrays.sort(arr, (a, b) -> b.val - a.val);
            memo = new long[n][n];
            for (long[] x : memo)
                Arrays.fill(x, -1);
            pw.println(dp(0, n - 1));
        }

        private long dp(int l, int r) {
            if (l > r)
                return 0;
            if (memo[l][r] != -1)
                return memo[l][r];
            long takeL = 1l * arr[l + n - 1 - r].val * Math.abs(arr[l + n - 1 - r].idx - l) + dp(l + 1, r);
            long takeR = 1l * arr[l + n - 1 - r].val * Math.abs(arr[l + n - 1 - r].idx - r) + dp(l, r - 1);
            return memo[l][r] = Math.max(takeL, takeR);
        }

        class pair {
            int val;
            int idx;

            public pair(int val, int idx) {
                this.val = val;
                this.idx = idx;
            }

            public String toString() {
                return val + " " + idx;
            }

        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}