Sum of special numbers

You are given an array of N numbers. Your task is to return the sum of all special numbers present in it.

A number is said to be special if the count of 0’s is equal to the count of 1's in its binary representation.

Note : We only consider 0’s to the right of Most Significant Bit while calculating the count of 0’s in a number’s binary representation, ie: 25=(000011001)₂ has 3 bits equal to 1 and 2 bits equal to 0.

Input Format :

The first line contains a single integer ‘T’ denoting the number of test cases to run. Then the test cases follow.

The first line contains a single integer N, where N represents the size of the array.

The next line contains N integers, denoting the elements of the given array.

Output Format :

For each test case print a single integer denoting the sum of special numbers in the array.

Output for each test case will be printed in a separate line,

Note :

You are not required to print anything; it has already been taken care of. Just implement the function.

Constraints :

1 <= T <= 10      
1 <= N <= 2 * 10^4
1 <= a[i] <= 10^9
Time limit: 1 sec

Sample Input 1 :

2
6
1 2 3 4 5 6
4
4 8 16 32

Sample Output 1 :

2
0

Explanation of sample input 1 :

For test case 1:
The binary representation of 2 is (10)2 it has an equal number of bits with values 1 and value 0.
Rest all the numbers have unequal count of bits with values 0 and 1.
Hence the final sum is 0+2+0+0+0+0 = 2

For test case 2:
None of the array elements has an equal count of bits with values 0 and 1.
4=(100) , 8=(1000) , 16=(10000) ,32=(100000)

Sample Input 2 :

2
2
12 9 
1
12

Sample Output 2 :

 21
 12