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: Accepted)

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

int N; pair<int, int> P[2020];
long long D[2020][2020];

int main()
{
	scanf ("%d", &N);
	for (int i = 0; i < N; i++) scanf ("%d", &P[i].first), P[i].second = i;
	sort(P, P + N);
	reverse(P, P + N);

	for (int i = 0; i < N; i++) for (int j = 0; i + j < N; j++){
		int k = i + j;
		D[i + 1][j] = max(D[i + 1][j], D[i][j] + 1ll * P[k].first * abs(P[k].second - i));
		D[i][j + 1] = max(D[i][j + 1], D[i][j] + 1ll * P[k].first * abs(P[k].second - (N - 1 - j)));
	}

	long long a = 0;
	for (int i = 0; i <= N; i++){
		a = max(a, D[i][N - i]);
	}
	printf ("%lld\n", a);

	return 0;
}