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;
typedef long long ll;
#define rep(i, n) for(ll i = 0, i##_len = (n); i < i##_len; i++)
#define reps(i, s, n) for(ll i = (s), i##_len = (n); i < i##_len; i++)
#define rrep(i, n) for(ll i = (n) - 1; i >= 0; i--)
#define rreps(i, e, n) for(ll i = (n) - 1; i >= (e); i--)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((ll)(x).size())
#define len(x) ((ll)(x).length())
#define endl "\n"

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    // ifstream in("input.txt");
    // cin.rdbuf(in.rdbuf());
    const ll inf = LONG_LONG_MIN / 2 + 1;
    ll n;
    cin >> n;
    vector<ll> a(n);
    rep(i, n) cin >> a[i];
    vector<pair<ll, ll>> pa(n);
    rep(i, n) pa[i] = make_pair(a[i], i);
    sort(all(pa), greater<pair<ll, ll>>());
    // dp[見た数][左においた数]
    vector<vector<ll>> dp(n + 1, vector<ll>(n + 1, inf));
    dp[0][0] = 0;
    rep(i, n) {
        ll val = pa[i].first;
        ll idx = pa[i].second;
        rep(j, n) {
            if (dp[i][j] == inf) continue;
            ll left = j, right = n - 1 - (i - j);
            dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + abs(right - idx) * val);
            dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j] + abs(left - idx) * val);
        }
    }
    ll ans = inf;
    rep(i, n + 1) ans = max(ans, dp[n][i]);
    cout << ans << endl;
    return 0;
}