human
You must lift the dam. With a lever. I will give it to you.

You must block the canal. With a rock. I will not give the rock to you." 

Danik urgently needs rock and lever! Obviously, the easiest way to get these things is to ask Hermit Lizard for them.

Hermit Lizard agreed to give Danik the lever. But to get a stone, Danik needs to solve the following task.

You are given a positive integer n, and an array a of positive integers. The task is to calculate the number of such pairs (i,j) that i<j and a_i \& a_j ≥ a_i ⊕ a_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND), and ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

Danik has solved this task. But can you solve it?

Input

Each test contains multiple test cases.

The first line contains one positive integer t (1 ≤ t ≤ 10) denoting the number of test cases. Description of the test cases follows.

The first line of each test case contains one positive integer n (1 ≤ n ≤ 10^5) — length of the array.

The second line contains n positive integers a_i (1 ≤ a_i ≤ 10^9) — elements of the array.

It is guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For every test case print one non-negative integer — the answer to the problem.

Example

Input

5
5
1 4 3 7 10
3
1 1 1
4
6 2 5 3
2
2 4
1
1

Output

1
3
2
0
0

Note

In the first test case there is only one pair: (4,7): for it 4 \& 7 = 4, and 4 ⊕ 7 = 3.

In the second test case all pairs are good.

In the third test case there are two pairs: (6,5) and (2,3).

In the fourth test case there are no good pairs.
Solve the task in PYTHON3.
 Write me a code to you must lift the dam?
assistant
#Here is code to you must lift the dam.

import itertools
import math
import sys
from collections import defaultdict

 
def stdinWrapper():
    data = '''1
0
3 3 3 3 3 

'''
   
    
    for line in data.split('\n'):
        yield line
 
if '--debug' not in sys.argv:
    def stdinWrapper():
        while True:
            yield input()
 
inputs = stdinWrapper()
 
def inputWrapper():
    return next(inputs)
 
def getType(_type):
    return _type(inputWrapper())
 
def getArray(_type):
    return [_type(x) for x in inputWrapper().split()]
 
 
 
''' Solution '''

def solve(a):

    def max_pow_of_2(x):
        return (pow(2, math.floor(math.log2(x))))


    a = list(reversed(sorted(a)))
    tgt = max_pow_of_2(a[0])
    bucket = 1
    result = 0
    for i in range(1, len(a)):
        # print(a[i], tgt, bucket, result)
        if a[i] >= tgt:
            result += bucket
            bucket += 1
        else:
            # print('reduce')
            tgt = max_pow_of_2(a[i])
            bucket = 1

    return result 
   
 
t = getType(int)

for _ in range(t):
    n = getType(int)
    a = getArray(int)
    res = solve(a)
    print(res)