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;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define ll long long

const long long MOD = 1000000007;

int main(){
  int n;
  cin >> n;
  vector<pair<ll,ll>> a(n);
  rep(i,n){
    int x;
    cin >> x;
    a.at(i) = make_pair(x,i);
  }
  sort(a.begin(),a.end());
  reverse(a.begin(),a.end());
  vector<bool> c(n,true);
  ll ans=0;
  rep(i,n){
  	ll m=0,t=0;
    int x=a.at(i).second;
    rep(j,n){
      if(c.at(j) && m < max(x,j) - min(x,j)){
        m=max(x,j)-min(x,j);
        t=j;
      }
    }
  	c.at(t)=false;
    ans+=m*a.at(i).first;
  }
  cout << ans << endl;
}