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 <stdlib.h>

int main()
{
	int i, j, N;
	long *A;
	int *toIndex;
	int *isFullFrom;
	int *isFullTo;
	long maxValue, tempMaxValue;
	int maxIndex;
	int orgIndex;
	int count = 0;
	long sum = 0;
	
	scanf("%d", &N);
	A = (long *) malloc(sizeof(long)*N);
	for (i = 0; i < N; i++)
		scanf(" %d", &A[i]);

	toIndex = (int *)malloc(sizeof(int)*N);
	isFullFrom = (int *)malloc(sizeof(int)*N);
	isFullTo = (int *)malloc(sizeof(int)*N);

	for (i = 0; i < N; i++) {
		isFullFrom[i] = 0;
        isFullTo[i] = 0;
		toIndex[i] = -1;
	}

	do {
		maxValue = -1;
		maxIndex = -1;
		orgIndex = -1;

		for (i = 0; i < N; i++) {
			if(isFullFrom[i]) continue;
			for (j = 0; j < N; j++) {
				if (isFullTo[j]) continue;
				tempMaxValue = A[i]*abs(i-j);
				if (maxValue < tempMaxValue) {
					orgIndex = i;
					maxIndex = j;
					maxValue = tempMaxValue;
				}
			}
		}
		
		toIndex[orgIndex] = maxIndex;
		isFullFrom[orgIndex] = 1;
		isFullTo[maxIndex] = 1;
		count++;
	} while (count < N);
	
	for (i = 0;i < N; i++) {
		sum += A[i]*abs(i - toIndex[i]);
	}
	printf("%ld\n",sum);
	fflush(stdout);
	free(A);
	return EXIT_SUCCESS;
}