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;
long long DP[2020][2020];
pair<long long,int> C[2020];
int main()
{
	int N;
	cin>>N;
	for(int i=0;i<N;i++)
	{
		cin>>C[i].first;
		C[i].second=i;
	}
	sort(C,C+N);
	reverse(C,C+N);
	for(int i=0;i<N;i++)
		for(int j=0;j<=i;j++)
		{
			int k=i-j;
			DP[j+1][k]=max(DP[j+1][k],DP[j][k]+C[i].first*C[i].second-j);
			DP[j][k+1]=max(DP[j][k+1],DP[j][k]+C[i].first*(N-C[i].second-1-k));
		}
	long long Ans=0;
	for(int i=0;i<=N;i++)
		Ans=max(Ans,DP[i][N-i]);
	cout<<Ans<<endl;
	return 0;
}