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;
const int maxn = 1e6 + 10;
struct node
{
    int val, dis, id, st;
};
int a[maxn];

bool cmp(node x, node y)
{
    if (x.val * x.dis == y.val * y.dis) {
        return x.dis > y.dis;
    }
    else return x.val * x.dis > y.val * y.dis;
}

bool take[maxn], seat[maxn];
int main()
{
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<node> v;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        for (int j = 1; j <= n; j++)
            if (j != i)
                v.push_back(node{a[i], abs(j - i), i, j});
    }
    sort(v.begin(), v.end(), cmp);
    long long ans = 0;
    for (int i = 0; i < v.size(); i++) {
        if (take[v[i].id] || seat[v[i].st]) continue;
        take[v[i].id] = true; seat[v[i].st] = true;
        ans += 1ll * v[i].dis * v[i].val;
    }
    cout << ans << '\n';
}