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 C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <deque>
using namespace std;

typedef long long ll;

deque<pair<ll, ll> > v;
deque<ll> L;
ll dp[2222][2222], n;

ll solve(ll s, ll e, ll val)
{
	ll cnt = s - 1 + n - e;

	if (s == e)
		return val + v[cnt].first * abs(s - v[cnt].second);
	else
		return dp[s][e] = max(dp[s][e], max(solve(s + 1, e, val + v[cnt].first * abs(v[cnt].second - s)), solve(s, e - 1, val + v[cnt].first * abs(v[cnt].second - e))));
}

int main(void)
{
	ll a, i;

	scanf("%lld", &n);
	for (i = 1; i <= n; i++)
	{
		scanf("%lld", &a);
		v.push_back(make_pair(a, i));
	}

	sort(v.begin(), v.end(), greater<>());

	printf("%lld", solve(1, n, 0));
	return 0;
}