This time, there will be no unnecessary information in the problem statement.
Given an array(A) of N integers, find the probability that sum of any selected triplet is a cube of an integer number.
A triplet is a set of any three indices. Two triplets with same indices but in different order are considered same (Ex: triplets with indices {2,4,5} and {2,5,4} are same).
Cube of a number x is x * x * x.
Input:
First line has an integer T=number of test cases.
Each test case has two lines.
First line of test case has an integer N=numbers in the array.
Second line of test case has an array of N integers.
Output:
Print answer for each test case in a separate lines with exactly 9 decimal digits.
Constraints:
1 ≤ T ≤ 40
3 ≤ N ≤ 500
1 ≤ Ai ≤ 2000
Case #1 : No triplet sum could be cube of an integer.
Case #2 : We can form 4 triplets.
Indices Values Sum
{1,2,3} {1,1, 6} 1+1+6 = 8
{1,2,4} {1,1,25} 1+1+25 = 27
{1,3,4} {1,6,25} 1+6+25 = 32
{2,3,4} {1,6,25} 1+6+25 = 32
Out of 4 resulting sums, 2 are cube of a number and therefore probability is 2/4.
