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: Wrong Answer)

import java.util.*;

public class Main {
    public static int abs(int a){
        return (a>0)?a:(-a);
    }
    
    public static void main(String[] args) throws Exception {
        Scanner sc=new Scanner(System.in);
        int N=sc.nextInt();
        long[] A=new long[N];
        boolean[] isSelected=new boolean[N];
        long[] leftmove=new long[N];
        long[] rightmove=new long[N];
        
        for(int i=0;i<N;i++){
            A[i]=sc.nextLong();
            isSelected[i]=false;
            leftmove[i]=A[i]*i;
            rightmove[i]=A[i]*(N-1-i);
        }
        
        int leftidx=0;
        int rightidx=N-1;
        long ans=0;
        
        for(int i=0;i<N;i++){
            long leftmax=-1;
            int leftmaxidx=0;
            long rightmax=-1;
            int rightmaxidx=0;
            for(int j=0;j<N;j++){
                if(leftmax<leftmove[j]&&!isSelected[j]){
                    leftmax=leftmove[j];
                    leftmaxidx=j;
                }
                if(rightmax<rightmove[j]&&!isSelected[j]){
                    rightmax=rightmove[j];
                    rightmaxidx=j;
                }
            }
            if(rightmax>leftmax){
                ans+=rightmax;
                rightidx--;
                isSelected[rightmaxidx]=true;
                for(int j=0;j<N;j++){
                    rightmove[j]=A[j]*abs(rightidx-j);
                }
            }else{
                ans+=leftmax;
                leftidx++;
                isSelected[leftmaxidx]=true;
                for(int j=0;j<N;j++){
                    leftmove[j]=A[j]*abs(j-leftidx);
                }
            }
        }
        System.out.println(ans);
    }
}