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, start, count) for(int i=(start); i<(int)(count); ++i)
#define rep(i, count) REP(i, 0, count)
#define ALLOF(c) (c).begin(), (c).end()

typedef long long ll;
typedef unsigned long long ull;

int N;
vector<ll> A;

struct infant {
    int i;
    ll v;
};

int main(void) {
    cin >> N;
    A.assign(N, 0);
    vector<infant> AR(N);
    vector<infant> AL(N);
    rep(i, N) {
        cin >> A[i];
        AL[i].i = i;
        AL[i].v = A[i] * i;
        AR[i].i = i;
        AR[i].v = A[i] * (N - i - 1);
    }

    sort(ALLOF(AL), [](const infant& lh, const infant& rh){ return lh.v > rh.v; });
    sort(ALLOF(AR), [](const infant& lh, const infant& rh){ return lh.v > rh.v; });

    ll ans = 0;
    ll li = 0, lc = 0;
    ll ri = 0, rc = 0;
    set<int> used;
    for(int i = 0; i < N; ++i) {

        infant lmax, rmax;
        for(int k = li; k < N; ++k) {
            if (used.find(AL[k].i) == used.end()) {
                lmax = AL[k];
                li = k;
                break;
            }
        }

        for (int k = ri; k < N; ++k)
        {
            if (used.find(AR[k].i) == used.end())
            {
                rmax = AR[k];
                ri = k;
                break;
            }
        }

        ll lv = lmax.v -(lc * A[lmax.i]);
        ll rv = rmax.v -(rc * A[rmax.i]);
        if (lv >= rv) {
            ans += lv;
            lc++;
            li++;
            used.insert(lmax.i);
        }
        else {
            ans += rv;
            rc++;
            ri++;
            used.insert(rmax.i);
        }
    }

    cout << ans << endl;

    return 0;
}