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: Time Limit Exceeded)

import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner stdIn = new Scanner(System.in);

        int n = stdIn.nextInt();
        long[] a = new long[n];
        for(int i=0; i<n; i++){
            a[i] = stdIn.nextLong();
        }

        int[] seed = new int[n];
        for(int i=0; i<n; i++){
            seed[i] = i;
        }
        Permutation p = new Permutation(seed);

        long max = 0;
        do{
            int[] perm = p.getPerm();
            long temp = 0;
            for(int i=0; i<perm.length; i++){
                temp += (Math.abs(perm[i] - i) * a[i]);
            }
            if(temp > max){
                max = temp;
            }
        }while(p.next());

        System.out.println(max);
    }
}

class Permutation{
    int[] seed;
    int[] perm;

    public Permutation(int [] seed){
        this.seed = new int[seed.length];
        for(int i=0; i<this.seed.length; i++){
            this.seed[i] = seed[i];
        }

        this.perm = new int[seed.length];
        for(int i=0; i<this.perm.length; i++){
            this.perm[i] = seed[i];
        }
        sort(this.perm, 0, this.perm.length);
    }

    public int[] getPerm(){
        return this.perm;
    }

    public boolean next(){
        for(int i=perm.length-1; i>0; i--){
            if(perm[i] > perm[i-1]){
                changePerm(i);
                return true;
            }
        }

        return false;
    }

    private void changePerm(int index){
        int targetIndex = index;
        int min = Integer.MAX_VALUE;
        for(int i=index; i<perm.length; i++){
            if(min > perm[i] && perm[index-1] < perm[i]){
                min = perm[i];
                targetIndex = i;
            }
        }

        int temp = perm[index - 1];
        perm[index - 1] = perm[targetIndex];
        perm[targetIndex] = temp;

        sort(perm, index, perm.length);
    }

    private void sort(int[] a, int from, int to){
        for(int i=from; i<to; i++){
            int x = a[i];
            int j = i;
            while(j > from && a[j-1] > x){
                a[j] = a[j-1];
                j--;
            }
            perm[j] = x;
        }
    }
}