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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

long long N;
pair<long long, int> A[1 << 18];
long long dp[2009][2009];

int main() {
	cin >> N;
	for (int i = 0; i < N; i++) {
		cin >> A[i].first;
		A[i].second = i + 1;
	}
	sort(A, A + N);

	for (int i = N - 1; i >= 0; i--) {
		for (int j = 1; j <= N - i; j++) {
			long long cl = j, cr = j + i;
			dp[cl + 1][cr] = max(dp[cl + 1][cr], dp[cl][cr] + A[i].first * (A[i].second - cl));
			dp[cl][cr - 1] = max(dp[cl][cr - 1], dp[cl][cr] + A[i].first * (cr - A[i].second));
		}
	}

	long long ret = 0;
	for (int i = 1; i <= N + 1; i++) ret = max(ret, dp[i][i - 1]);
	cout << ret << endl;
	return 0;
}