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.Scanner;
import java.util.TreeSet;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int a[] = new int[n];
		for(int i=0;i<n;i++) {
			a[i] = sc.nextInt();
		}
		TreeSet<Integer> ts = new TreeSet<>(new Comparator<Integer>() {

			@Override
			public int compare(Integer o1, Integer o2) {
				// TODO Auto-generated method stub
				if(a[o1]==a[o2]) {
					return o1-o2;
				}
				return a[o1]-a[o2];
			}
			
		});
		for(int i=0;i<n;i++) {
			ts.add(i);
		}
		int order[] = new int[n];
		for(int i=0;i<n;i++) {
			order[i] = ts.first();
			ts.remove(ts.first());
		}
		long dp[][] = new long[n][n];
		for(int len=0;len<n;len++) {
			for(int i=0;i<n;i++) {
				int j = i+len;
				if(j>n-1) continue;
				if(i==j) {
					dp[i][j] = (long) Math.abs(order[0]-i)*a[order[0]];
				}
				else {
					dp[i][j] = Math.max((long) Math.abs(order[len]-i)*a[order[len]]+dp[i+1][j],(long) Math.abs(order[len]-j)*a[order[len]]+dp[i][j-1]);
				}
			}
		}
		System.out.println(dp[0][n-1]);
	}
}