"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.
"""
                            pppppppppppppppppppp
                         ppppp  ppppppppppppppppppp
                      ppppppp    ppppppppppppppppppppp
                      pppppppp  pppppppppppppppppppppp
                      pppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppppp
       ppppppppppppppppppppppppppppppppppppppppppppppp  pppppppppppppppppppp
      pppppppppppppppppppppppppppppppppppppppppppppppp  ppppppppppppppppppppp
     ppppppppppppppppppppppppppppppppppppppppppppppppp  pppppppppppppppppppppp
    ppppppppppppppppppppppppppppppppppppppppppppppp    pppppppppppppppppppppppp
   pppppppppppppppppppppppppppppppppppppppppppppp     pppppppppppppppppppppppppp
  ppppppppppppppppppppppppppppppppppppppppppppp      pppppppppppppppppppppppppppp
  pppppppppppppppppppppppppppppppp               pppppppppppppppppppppppppppppppp
  pppppppppppppppppppppppppppp     pppppppppppppppppppppppppppppppppppppppppppppp
  ppppppppppppppppppppppppppp    pppppppppppppppppppppppppppppppppppppppppppppppp
    pppppppppppppppppppppppp  pppppppppppppppppppppppppppppppppppppppppppppppppp
     ppppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppppp
      pppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppp
       ppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppppp
                              pppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppp  pppppppp
                              ppppppppppppppppppppp    ppppppp
                                 ppppppppppppppppppp  ppppp
                                    pppppppppppppppppppp
"""


import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
from decimal import Decimal
# sys.setrecursionlimit(2 * (10 ** 5))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(str(var)+"\n")
def outa(*var, end="\n"): sys.stdout.write(' '.join(map(str, var)) + end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]


for _ in range(int(data())):
    n = int(data())
    arr = l()
    dp = [0] * 33
    for i in arr:
        for j in range(32, -1, -1):
            if i & (1 << j):
                dp[j] += 1
                break
    answer = 0
    for i in range(32, -1, -1):
        answer += (dp[i] * (dp[i] - 1)) // 2
    out(answer)
