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;
using ll = long long;

#define ALL(obj) (obj).begin(),(obj).end()
#define SPEED cin.tie(0);ios::sync_with_stdio(false);

template<class T> using PQ = priority_queue<T>;
template<class T> using PQR = priority_queue<T,vector<T>,greater<T>>;

constexpr long long MOD = (long long)1e9 + 7;
constexpr long long MOD2 = 998244353;
constexpr long long HIGHINF = (long long)1e18;
constexpr long long LOWINF = (long long)1e15;
constexpr long double PI = 3.1415926535897932384626433L;

template <class T> vector<T> multivector(size_t N,T init){return vector<T>(N,init);}
template <class... T> auto multivector(size_t N,T... t){return vector<decltype(multivector(t...))>(N,multivector(t...));}
template <class T> void corner(bool flg, T hoge) {if (flg) {cout << hoge << endl; exit(0);}}
template <class T, class U>ostream &operator<<(ostream &o, const map<T, U>&obj) {o << "{"; for (auto &x : obj) o << " {" << x.first << " : " << x.second << "}" << ","; o << " }"; return o;}
template <class T>ostream &operator<<(ostream &o, const set<T>&obj) {o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr) o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;}
template <class T>ostream &operator<<(ostream &o, const multiset<T>&obj) {o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr) o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;}
template <class T>ostream &operator<<(ostream &o, const vector<T>&obj) {o << "{"; for (int i = 0; i < (int)obj.size(); ++i)o << (i > 0 ? ", " : "") << obj[i]; o << "}"; return o;}
template <class T, class U>ostream &operator<<(ostream &o, const pair<T, U>&obj) {o << "{" << obj.first << ", " << obj.second << "}"; return o;}
template <template <class tmp>  class T, class U> ostream &operator<<(ostream &o, const T<U> &obj) {o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr)o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;}
void print(void) {cout << endl;}
template <class Head> void print(Head&& head) {cout << head;print();}
template <class Head, class... Tail> void print(Head&& head, Tail&&... tail) {cout << head << " ";print(forward<Tail>(tail)...);}
template <class T> void chmax(T& a, const T b){a=max(a,b);}
template <class T> void chmin(T& a, const T b){a=min(a,b);}
void YN(bool flg) {cout << (flg ? "YES" : "NO") << endl;}
void Yn(bool flg) {cout << (flg ? "Yes" : "No") << endl;}
void yn(bool flg) {cout << (flg ? "yes" : "no") << endl;}

int main() {
	int N; cin >> N;
	vector<ll> A(N+1);
	for(int i = 1; i <= N;++i) cin >> A[i];
	auto dp = multivector(N+1,N+1,-1LL);
	dp[0][0] = 0;
	vector<pair<ll,ll>> ord(N+1);
	for(int i = 1; i <= N; ++i) {
		ord[i] = {A[i],i};
	}
	sort(ALL(ord),greater<>());
	for(int i = 0; i < N; ++i) {
		ll x = ord[i].first;
		ll idx = ord[i].second;
		int sum = i;
		for(int nl = 0; nl <= sum; ++nl) {
			int nr = sum - nl;
			int l = nl+1;
			int r = N - nr;
			if(nl+1<=N&&dp[nl][nr]!=-1) chmax(dp[nl+1][nr],dp[nl][nr]+abs(idx-l)*x);
			if(nr+1<=N&&dp[nl][nr]!=-1) chmax(dp[nl][nr+1],dp[nl][nr]+abs(idx-r)*x);
		}
	}
	ll ans = 0;
	for(int i = 0; i <= N; ++i) {
		chmax(ans,dp[i][N-i]);
	}
	cout << ans << endl;

    return 0;
}