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 <stdio.h>
#include <math.h>
#include <stdlib.h>
typedef long long ll;
int compare_int(const void* a, const void* b){
		return *(int*)b - *(int*)a;
}

int LorR(ll left, ll right, ll n) {
	if (right +left >= n*2) { return 1; }
	else { return 0; }//遠い方を指す
}

int main() {
	ll i, n, l, r, a[2020][2];//活発度,元々いた座標
	ll ans=0;
	for (i = 0; i < 2020; i++) { a[i][0] = 0; a[i][1] = 0; }
	scanf("%lld", &n);
	for (i = 1; i <= n; i++) {
		a[i][1] = i;
		scanf("%lld", &a[i][0]);
	}
	qsort(a, n, sizeof(a[0]), compare_int);
	l = 0; r = n+1;
	for (i = 1; i <= n; i++) {
		if (LorR(l, r, a[i][1]) == 1) {
			ans += a[i][0] * (r - 1 - a[i][1]);
			r--;
		}
		else {
			ans += a[i][0] * (a[i][1] - l + 1);
			l++;
		}
	}
	printf("20", ans);
}