Given an array of n integers, count all different triplets whose sum is equal to the perfect cube i.e, for any i, j, k(i < j < k) satisfying the condition that a[i] + a[j] + a[j] = X3 where X is any integer. 3 ≤ n ≤ 1000, 1 ≤ a[i, j, k] ≤ 5000

Input:

N = 5

2 5 1 20 6

Output:

3

Explanation:

There are only 3 triplets whose total sum is a perfect cube.

Indices Values SUM

0 1 2 2 5 1 8

0 1 3 2 5 20 27

2 3 4 1 20 6 27

Since 8 and 27 are prefect cube of 2 and 3.