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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef pair<long long, int> pli;

int main(){
	ios::sync_with_stdio(0);
	cout << setprecision(10) << fixed;
	int n;
	cin >> n;
	vector<pli> a(n + 1);
	for (int i = 1; i <= n; ++i){
		cin >> a[i].first;
		a[i].second = i + 1;
	}
	sort(a.begin() + 1, a.end());
	reverse(a.begin() + 1, a.end());
	vector<vector<long long>> dp(n + 2, vector<long long>(n + 2, 0));
	for (int i = 1; i <= n; ++i){
		for (int j = 0; j <= i; ++j){
			long long x = a[i].first;
			long long l = abs(a[i].second - (i - j));
			long long r = abs((n - j) - a[i].second);
			dp[i][j] = max(dp[i][j], dp[i - 1][j] + x * l);
			dp[i][j + 1] = max(dp[i][j + 1], dp[i - 1][j] + x * r);
		}
	}
	long long ans = 0;
	for (int i = 0; i <= n; ++i){
		ans = max(ans, dp[n][i]);
	}
	cout << ans;
	return 0;
}