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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import kotlin.math.max

fun main() {
    val n = readLine()!!.toInt()
    val a = readLine()!!.split(" ").map { it.toLong() }

    var r = 0
    var l = n - 1

    val done = MutableList(n) { false }
    var ans = 0L

    repeat(n) {
        val p = (0 until n).filter { !done[it] }
            .maxBy {
                a[it] * max(l - it, it - r)
            }!!

        done[p] = true;
        ans += a[p] * max(l - p, p - r)

        when {
            l - p > p - r -> l--
            else -> r++
        }
    }

    println(ans)
}