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 rep(i, a, b) for (int i = (a), i##end = (b); i <= i##end; ++i)
#define per(i, a, b) for (int i = (a), i##end = (b); i >= i##end; --i)

mt19937 Rnd(chrono::high_resolution_clock::now().time_since_epoch().count());

template <typename T>
inline void chkmax(T &x, T y) { if (x < y) x = y; }

template <typename T>
inline void chkmin(T &x, T y) { if (x > y) x = y; }

inline int read() {
  #define nc getchar()
  int x = 0;
  char c = nc;
  while (c < 48) c = nc;
  while (c > 47) x = x * 10 + (c ^ 48), c = nc;
  return x;
  #undef nc
}

typedef long long ll;
typedef pair <int, int> pii;
const int maxn = 2010;
int n;
pii A[maxn];

ll dp[2][maxn];

int getdis(int x) {
  return max(x, n - x + 1);
}

int get(int x, int y) {
  return x < y ? y - x : x - y;
}

int main() {
  cin >> n;
  rep(i, 1, n) A[i] = pii(read(), i);
  sort(A + 1, A + n + 1, [] (pii P, pii Q) {
    return P.first == Q.first ? getdis(P.second) > getdis(Q.second) : P.first > Q.first;
  });
  int sel = 1;
  rep(i, 1, n) {
    rep(j, 0, i) {
      dp[sel][j] = 0;
      if (j < i) chkmax(dp[sel][j], dp[sel ^ 1][j] + 1ll * A[i].first * get(n - (i - j) + 1, A[i].second));
      if (j) chkmax(dp[sel][j], dp[sel ^ 1][j - 1] + 1ll * A[i].first * get(j, A[i].second));
    }
    sel ^= 1;
  }
  ll res = 0;
  rep(i, 0, n) chkmax(res, dp[sel ^ 1][i]);
  cout << res << endl;
  return 0;
}