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;
vector<pair<int, int> > a(2010);
long dp[2010][2010];
template<class T> inline bool chmax(T& a, const T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}
int main() {
	int n, x;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> x;
		a.push_back(make_pair(x, i));
	}
	sort(a.begin(), a.end(), greater<pair<int, int> >());
	long ans = 0;
	for (int l = 0; l < n; l++) {
		for (int r = 0; r < n; r++) {
			if (l + r == n) {
				chmax(ans, dp[l][r]);   break;
			}
			else {
				chmax(dp[l + 1][r], dp[l][r] + (long)a[l + r].first * abs(l - a[l + r].second));
				chmax(dp[l][r + 1], dp[l][r] + (long)a[l + r].first * abs((n - 1 - r) - a[l + r].second));
			}
		}
	}
	cout << ans;
}