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>

#include <boost/range/adaptors.hpp>
#include <boost/range/irange.hpp>

using namespace std;
using namespace boost;
using namespace boost::adaptors;

int main() {
    int64_t n;
    cin >> n;

    vector<int64_t> a(n);
    priority_queue<pair<int64_t, int64_t>> q;
    for (auto&& aa : a | indexed()) {
        cin >> aa.value();
        q.emplace(aa.value(), aa.index());
    }

    vector dp(n + 1, -1L);
    dp[0] = 0;
    int64_t r = 0;
    while (!q.empty()) {
        auto t = q.top();
        q.pop();

        vector next(n + 1, -1L);
        for (auto i : irange(0L, n)) {
            if (dp[i] < 0) {
                continue;
            }

            next[i + 1] = max(next[i + 1], dp[i] + t.first * abs(t.second - i));
            next[i] = max(next[i],
                          dp[i] + t.first * abs(t.second - (n - r + i - 1)));
        }

        ++r;
        dp = std::move(next);
    }

    cout << *max_element(dp.begin(), dp.end()) << endl;
}