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.Arrays;
import java.util.Scanner;

class Main {
    static Long[][][] dp;
    static int INF = (int) 1e9 + 7;
    static Integer[][] arr;
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int N = s.nextInt();
        arr = new Integer[N][2];
        for(int i = 0; i < N; i++) {
            arr[i][0] = s.nextInt();
            arr[i][1] = i;
        }
        Arrays.sort(arr, (a, b) -> b[0] - a[0]);
        dp = new Long[N][N][N];
        System.out.println(dfs(0, 0, N-1));
    }

    private static long dfs(int index, int left, int right) {
        if(index == arr.length) {
            return 0;
        }
        if(dp[index][left][right] != null) {
            return dp[index][left][right];
        }
        return dp[index][left][right] = Math.max(dfs(index+1, left+1, right) + (long)arr[index][0]*Math.abs(arr[index][1] - left),
                dfs(index+1, left, right-1) + (long)arr[index][0]*Math.abs(arr[index][1] - right));
    }
}