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;

template<class T>
void chmax(T &x, T y) { if (x < y) x = y; }

const int N = 2010;
ll dp[N][N];

int main() {
    int n;
    cin >> n;
    vector<pair<ll, int>> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i].first;
        a[i].second = i;
    }
    sort(a.rbegin(), a.rend());
    ll ans = 0;
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= n - i; j++) {
            if (!i && !j) continue;
            if (i) chmax(dp[i][j], dp[i - 1][j] + a[i + j - 1].first * (a[i + j - 1].second - i + 1));
            if (j) chmax(dp[i][j], dp[i][j - 1] + a[i + j - 1].first * (n - j - a[i + j - 1].second));
            if (i + j == n) chmax(ans, dp[i][j]);
        }
    }
    cout << ans << "\n";
}