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;
typedef long long ll;
typedef pair<int, int> P;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repr(i, l, r) for (int i = l; i <= (int)(r); i++)
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define MOD (int) (1e9+7)
#define INF (int) 2e9
#define LLINF (ll) 2e18

ll dp[2005][2005];

int main(){
    int n;
    cin >> n;
    P p[n];
    rep(i, n){
        int a;
        cin >> a;
        p[i] = P(a, i);
    }

    sort(p, p+n, greater<>());

    dp[0][0] = 0;

    rep(xi, n){
        rep(i, xi+1){
            dp[xi-i+1][i] = max(dp[xi-i+1][i], dp[xi-i][i] + p[xi].first * abs(p[xi].second - (xi-i)));
            dp[xi-i][i+1] = max(dp[xi-i][i+1], dp[xi-i][i] + p[xi].first * abs(p[xi].second - (n-i-1)));
        }
    }
    
    ll ans = 0;
    rep(i, n+1) chmax(ans, dp[n-i][i]);

    cout << ans << endl;
    return 0;
}