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 int long long
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rep1(i,n) for(int i=1;i<=(int)(n);++i)
#define rep11(i,n) for(int i=1;i<(int)(n);++i)
#define repo(i,o,n) for(int i=o;i<(int)(n);++i)
#define repm(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define pb(n) push_back(n)
#define mp make_pair
#define INF LONG_LONG_MAX
#define MOD 1000000007
#define Yes cout << "Yes" << endl;
#define No cout << "No" << endl;

int n,ans;
bool f[2000],F[2000];
priority_queue<pair<int,int>> pq[2000];

signed main() {
    cin >> n;
    rep(i,n) {
        int a;
        cin >> a;
        rep(j,n) pq[i].push(mp(a*abs(i-j), j));
    }
    rep(i,n) {
        int idx,now = 0;
        pair<int,int> m;
        rep(j,n-1) {
            if (!F[j]) {
                while (f[pq[j].top().se]) pq[j].pop();
                auto fir = pq[j].top();
                pq[j].pop();
                while(f[pq[j].top().se]) pq[j].pop();
                auto sec = pq[j].top();
                pq[j].push(fir);
                if (now < fir.fi) {
                    now = fir.fi;
                    idx = j;
                    m = fir;
                }
            }
        }
        ans += m.fi;
        f[m.se] = true;
        F[idx] = true;
    }
    rep(i,n) if (!F[i]) {
        while(f[pq[i].top().se]) pq[i].pop();
        ans += pq[i].top().fi;
    }
    cout << ans << endl;
}