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>
using namespace std;
#define ll long long

int main() {
    int N;
    cin >> N;
    vector<pair<ll, int>> A(N);
    for (int i = 0; i < N; i++) {
        cin >> A[i].first;
        A[i].second = i;
    }
    sort(A.begin(), A.end());
    reverse(A.begin(), A.end());
    vector<vector<ll>> dp(N+1, vector<ll>(N+1, 0));
    for (int i = 0; i < N; i++) {
        dp[0][i+1] = dp[0][i] + abs(N-1-i-A[i].second) * A[i].first;
        dp[i+1][0] = dp[i][0] + abs(A[i].second-i) * A[i].first;
        for (int j = 1; j <= i; j++) {
            dp[j][i+1-j] = max(dp[j-1][i+1-j] + abs(j-1-A[i].second) * A[i].first, dp[j][i-j] + abs(N-i-1+j - A[i].second) * A[i].first);
        }
    }
    ll ans = 0;
    for (int i = 0; i <= N; i++) ans = max(ans, dp[i][N-i]);
    cout << ans << endl;
    return 0;
}