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: Time Limit Exceeded)

import java.lang.Math.abs

fun main() = ABC163E2.main()

object ABC163E2 {
    fun log(message: Any?) {
        println(message)// TODO: ←消す
    }

    val infants: List<Infant>

    init {
        val N = readInt()
        val A = readLongs()

        infants = A.mapIndexed { index, it -> Infant(pos = index, value = it) }
            .sortedBy { -it.value * N + it.pos }   // 価値=降順, 位置=昇順
    }

    data class Problem(
        val left: Int,  // 左に移動させた人数
        val right: Int  // 右に移動させた人数
    ) {
        val done get() = left + right   // 既に移動させた人数
        val remain get() = infants.size - done    // 残り人数
        val nextInfant get() = infants.getOrNull(done)
        val leftMoveValue: Long?
            get() = nextInfant?.let {
                it.value * abs(it.pos - left)
            }
        val rightMoveValue: Long?
            get() = nextInfant?.let {
                it.value * abs(infants.size - 1 - right - it.pos)
            }

        fun getOneChoiceOrNull(): Long? {
            return when {
                remain == 0 -> 0L
                else -> null
            }
        }
    }

    data class Infant(val pos: Int, val value: Long)

    fun main() {

        val table = LazyMap<Problem, Long> { problem ->
            problem.getOneChoiceOrNull() ?: problem.let {
                // subproblems
                val subProblems = listOf(
                    Problem(left = it.left + 1, right = it.right),
                    Problem(left = it.left, right = it.right + 1)
                ).filter { it.left >= 0 && it.right >= 0 }
                // this value
                val value = subProblems.map { sub ->
                    get(sub) + if (it.left < sub.left) it.leftMoveValue!! else it.rightMoveValue!!
                }.max()!!
                value
            }
        }

        val ans = table[Problem(0, 0)]
        println(ans)
    }


    /*####################################################################################*/

    fun readInt() = readLine()!!.toInt()
    fun readLongs() = readLine()!!.split(" ").map { it.toLong() }
    fun <E> List<E>.dropAt(index: Int): List<E> {
        val buf = this.toMutableList()
        buf.removeAt(index)
        return buf
    }

    class LazyMap<K, V>(private val init: LazyMap<K, V>.(K) -> V) : Map<K, V> {
        private val memo = HashMap<K, V>()
        override fun get(key: K): V {
            return memo[key] ?: init(key).memorizingAt(key)
        }

        private fun V.memorizingAt(key: K): V {
            memo[key] = this
            return this
        }

        override val entries: Set<Map.Entry<K, V>>
            get() = memo.entries
        override val keys: Set<K>
            get() = memo.keys
        override val size: Int
            get() = memo.size
        override val values: Collection<V>
            get() = memo.values

        override fun containsKey(key: K): Boolean = memo.containsKey(key)
        override fun containsValue(value: V): Boolean = memo.containsValue(value)
        override fun isEmpty(): Boolean = memo.isEmpty()
    }
}