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;
int main(){
    long long n;
    cin >> n;
    pair<long long, long long>v[n + 3];
    for(long long i = 1; i <= n; i++){
        long long x;
        cin >> x;
        v[i] = {x, i};
    }
    sort(v + 1, v + n + 1, greater<pair<long long, long long> >());
    long long dp[n + 3][n + 3];
    dp[1][0] = v[1].first * abs(v[1].second - n);
    dp[1][1] = v[1].first * abs(v[1].second - 1);
    for(int i = 2; i <= n; i++){
        dp[i][0] = dp[i - 1][0] + v[i].first * abs(v[i].second - (n - i + 1));
        dp[i][i] = dp[i - 1][i - 1] + v[i].first * abs(v[i].second - i);
        for(int j = 1; j < i; j++){
            dp[i][j] = max(dp[i - 1][j] + v[i].first * abs(v[i].second - (n - i + j + 1)), dp[i - 1][j - 1] + v[i].first * abs(v[i].second - j));
        }
    }
    long long ans = 0;
    for(int i = 0; i <= n; i++){
        ans = max(ans, dp[n][i]);
    }
    cout << ans;
    return 0;
}