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 all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define MIN(a) *min_element(all(a))
#define MAX(a) *max_element(all(a))
#define SUM(a) accumulate(all(a), 0LL)
#define REP(i, n) for(int (i)=0; (i)<(n); (i)++)
#define RREP(i, n) for(int (i)=(n)-1; (i)>=0; (i)--)
#define FOR(i, m, n) for(int (i)=(m); (i)<(n); i++)
#define FORR(i, m, n) for(int (i)=(n)-1; (i)>=(m); i--)
#define debug(x) cerr << #x << " = " << x << endl
//#define int long long

typedef long long ll;
//ll const MOD = 998244353;
ll const MOD = 1e9+7;
int const inf = 1e9;
ll const INF = 1e18;

inline void ios_(){cin.tie(0); ios::sync_with_stdio(false);}
template<typename T> int size(const T& a){return (int)a.size();}
template<typename T> T Div(T a, T b){return (a + b - 1) / b;}
template<typename T> bool chmin(T& a, const T& b){if(a > b){a = b; return true;} return false;}
template<typename T> bool chmax(T& a, const T& b){if(a < b){a = b; return true;} return false;}



signed main(){
    int n;
    cin >> n;
    vector<ll> v(n);
    for(auto&& i : v) cin >> i;

    vector<int> order(n);
    REP(i, n) order[i] = i;
    sort(all(order), [&](auto a, auto b){return v[a] > v[b];});

    vector<bool> used(n, false);
    ll ans = 0;

    int left = 0, right = n-1;

    REP(i, n){
        ll score = -1;
        int p = 0;
        REP(j, n){
            if(used[j]) continue;
            ll l = abs(j - left) * v[j], r = abs(right - j) * v[j];
            if(chmax(score, max(l, r))){
                p = j;
            }
        }
        ll l = abs(p - left) * v[p], r = abs(right - p) * v[p];
        if(l > r){
            ans += l;
            left++;
        }else{
            ans += r;
            right--;
        }
        used[p] = true;
    }

    cout << ans << endl;

    return 0;
}