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)

fun main(args: Array<String>) {
    val n = readLine()!!.toInt()
    val aList = readLine()!!.split(' ').map(String::toLong).toMutableList()
    aList.add(0, 0)
    val apList = (1..n).map { Pair(it, aList[it]) }.sortedByDescending { it.second }


    var l = 1
    var r = n
    var ans = 0L
    for (p in apList) {
        val (i, a) = p
        val la = aList[l]
        val ra = aList[r]
        val ldiff = Math.abs(l - i).toLong()
        val rdiff = Math.abs(r - i).toLong()
        val lVal = ldiff * a + ldiff * la
        val rVal = rdiff * a + rdiff * ra
        if (lVal > rVal) {
            ans += ldiff * a
            l++
        } else {
            ans += rdiff * a
            r--
        }
    }
    println(ans)
}