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.util.Comparator;
import java.util.List;
import java.util.Scanner;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
import java.util.stream.LongStream;

public class Main {
    public static void main(final String[] args) {
        final Scanner scanner = new Scanner(System.in);
        final int n = scanner.nextInt();
        final List<Child> children = IntStream.rangeClosed(1, n)
                .mapToObj(i -> new Child(scanner.nextLong(), i))
                .sorted(Comparator.comparingLong(child -> -child.value))
                .collect(Collectors.toList());

        final long[][] dpTable = new long[n + 1][n + 1];
        for (int i = 1; i <= n; i++) {
            final Child child = children.get(i - 1);
            dpTable[i][0] = dpTable[i - 1][0] + child.value * (child.position - i);
            dpTable[0][i] = dpTable[0][i - 1] + child.value * (n - i + 1 - child.position);
            for (int j = 1; j < i; j++) {
                final int diff = i - j;
                dpTable[j][diff] = Math.max(dpTable[j - 1][diff] + child.value * (child.position - j), dpTable[j][diff - 1] + child.value * (n - diff + 1 - child.position));
            }
        }

        final long answer = LongStream.range(0, n)
                .map(i -> dpTable[(int) i][n - (int) i])
                .max().orElseThrow();
        System.out.println(answer);
    }

    static class Child {
        final long value;
        final long position;

        public Child(final long value, final long position) {
            this.value = value;
            this.position = position;
        }
    }
}