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: Time Limit Exceeded)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define INF 0x7fffffff
#define INFL 0x7fffffffffffffffLL

vector<pair<ll,ll>> a(2000);
ll n;
ll dp[2000][2000];

ll calc(ll x, ll y) {
	int idx = x + ((n-1) - y);
	if (idx == n) return 0;
	//if (dp[x][y]!=0) return dp[x][y];
	ll ans = max( abs(a[idx].second-x)*a[idx].first + calc(x+1 , y) , abs(a[idx].second-y)*a[idx].first + calc(x , y-1) );
	dp[x][y] = ans;
	return ans; 
}

int main() {
	ll		b,c,i,j,k,m,x,y,ans = 0;
	string	str;

	cin >> n;

	for(i=0;i<n;i++) {
		cin >> j;
		a[i] = make_pair(j , i);
	}
	sort(a.begin(), a.end());
	reverse(a.begin(), a.end());

	x=0;
	y=n-1;

	ans = calc(x , y);

	cout << ans << endl;

}