"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.
import sys
import math
 
# Constants
mod = 998244353
 
 
# Functions Settings
dynamic_factorial = False
 
 
 
# Pre-defined function
# Begin
 
# Testcase handler
def main(test_case = False):
    n = int(input()) if test_case else 1
    for _ in range(n):
        test()
        print()
 
# Binary method of power
def fast_pow(a:int, b:int):
    res = 1
    while b > 0:
        if b & 1:
            res *= a
        a *= a
        b >>= 1
    return res
 
# Combination of 2
def c2(n):
    return n * (n-1) // 2
 
# Get the list form of input
def get():
    return list(map(int, input().split()))
 
# Number of 1s in binary form
def bits(n: int):
    return list(bin(n)).count('1')
 
# For interactive problems
def flush():
    sys.stdout.flush()
 
# Print array and split it with space
def parr(arr):
    print(*arr, sep=' ')
 
# GCD (int a, int b)
# Return: value of GCD(a, b)
def gcd(a, b):
    while b:
        if b % a == 0:
            break
        tmp = a
        a = b % a
        b = tmp
    return a
 
# Extended GCD (int a, int b)
# Return: array of [GCD(a, b), [x0, y0]]
# Where a * x0 + b * y0 = 1
# If GCD(a, b) != 1 then return False
def ext_gcd(a: int, b: int):
    if (b == 0):
        return [a, [1, 0]]
 
    res = ext_gcd(b, a % b)
    g = res[0]
    if g != 1:
        return False
    x1 = res[1][0]
    y1 = res[1][1]
    x = y1
    y = x1 - y1 * (a // b)
 
    return [g, [x, y]]
 
# Get all different values of array a with their count
def get_set(a):
    res = {}
    for x in a:
        if x in res:
            res[x] += 1
        else:
            res[x] = 1
    return res
 
# Factorial
fact = {0: 1}
def factorial(n):
    if dynamic_factorial:
        if n not in fact:
            fact[n] = factorial(n-1) * n
        return fact[n]
    else:
        res = 1
        while n > 0:
            res = (res * n) % mod
            n -= 1
        return res
 
# Sum of array
def sum_of(a):
    res = 0
    for x in a:
        res = (res + x) % mod
    return res
 
# Prod of array
def prod_of(a):
    res = 1
    for x in a:
        res = (res * x) % mod
    return res
 
# End

def test():
    n = get()[0]
    a = get()
    b = [-1] * n
    for i in range(n):
        b[i] = len(bin(a[i])) - 3
    b.sort()
    # print(b)
    i = 0
    ans = 0
    while i < n:
        while i+1 < n and b[i] == b[i+1]:
            i += 1
        ans += c2(b.count(b[i]))
        i += 1
    print(ans)

main(test_case = True)