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"bits/stdc++.h"
using namespace std;
typedef long long ll;
#define int ll
const int N=2003;
int dp[N][N];
int ans;
void cmax(int &x,int y){x=max(x,y);}
int32_t main()
{
	#ifndef ONLINE_JUDGE
	freopen("in","r",stdin);
	freopen("out","w",stdout);
	#endif
	int n;
	cin>>n;
	pair<int,int> a[n+1];
	for(int i=1;i<=n;i++)
		cin>>a[i].first,a[i].second=i;
	sort(a+1,a+n+1);reverse(a+1,a+n+1);


	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<i;j++)
		{
			int l=j;
			int r=i-1-j;
			cmax(dp[l+1][r],dp[l][r]+abs(l+1-a[i].second)*a[i].first);
          	r=j;l=i-1-j;
			cmax(dp[l][r+1],dp[l][r]+abs(n-r-a[i].second)*a[i].first);

		}
	}
	for(int i=0;i<=n;i++)cmax(ans,dp[i][n-i]);
	cout<<ans<<"\n";
}