description: One day, early in the morning, you decided to buy yourself a bag of chips in the nearby store. The store has chips of n different flavors. A bag of the i-th flavor costs a_i burles.
  
  The store may run out of some flavors, so you'll decide which one to buy after arriving there. But there are two major flaws in this plan: 
  
    1. you have only coins of 1, 2 and 3 burles; 
    2. since it's morning, the store will ask you to pay in exact change, i. e. if you choose the i-th flavor, you'll have to pay exactly a_i burles. 
  
  
  
  Coins are heavy, so you'd like to take the least possible number of coins in total. That's why you are wondering: what is the minimum total number of coins you should take with you, so you can buy a bag of chips of any flavor in exact change?
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
  
  The first line of each test case contains the single integer n (1 ≤ n ≤ 100) — the number of flavors in the store.
  
  The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the cost of one bag of each flavor.
  
  Output
  
  For each test case, print one integer — the minimum number of coins you need to buy one bag of any flavor you'll choose in exact change.
  
  Example
  
  Input
  
  
  4
  1
  1337
  3
  10 8 10
  5
  1 2 3 4 5
  3
  7 77 777
  
  
  Output
  
  
  446
  4
  3
  260
  
  Note
  
  In the first test case, you should, for example, take with you 445 coins of value 3 and 1 coin of value 2. So, 1337 = 445 ⋅ 3 + 1 ⋅ 2.
  
  In the second test case, you should, for example, take 2 coins of value 3 and 2 coins of value 2. So you can pay either exactly 8 = 2 ⋅ 3 + 1 ⋅ 2 or 10 = 2 ⋅ 3 + 2 ⋅ 2.
  
  In the third test case, it's enough to take 1 coin of value 3 and 2 coins of value 1.
solution: 
  inf = 2000000000
  
  def solve1(a, n, n1, n2): # return n1 + n2 + minN3 required
      ans = 0
      for x in a:
          minCntsForX = inf
          for nn1 in range(n1 + 1):
              for nn2 in range(n2 + 1):
                  y = x - nn1 - nn2 * 2
                  if y >= 0 and y % 3 == 0:
                      minCntsForX = min(minCntsForX, n1 + n2 + y // 3)
          ans = max(ans, minCntsForX)
      return ans
                  
              
  
  def solve(a, n):
      ans = inf
      for n1 in range(5):
          for n2 in range(5): # guess that all ans has max of 10 n1 and 10 n2
              ans = min(ans, solve1(a, n, n1, n2))
      return ans
  
  def main():
      
      t = int(input())
      allans = []
      for _ in range(t):
          n = int(input())
          a = readIntArr()
          allans.append(solve(a, n))
      multiLineArrayPrint(allans)
      
      return
  
  
  import sys
  input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
  # input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
  
  def oneLineArrayPrint(arr):
      print(' '.join([str(x) for x in arr]))
  def multiLineArrayPrint(arr):
      print('\n'.join([str(x) for x in arr]))
  def multiLineArrayOfArraysPrint(arr):
      print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
   
  def readIntArr():
      return [int(x) for x in input().split()]
  # def readFloatArr():
  #     return [float(x) for x in input().split()]
   
  def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
      dv=defaultValFactory;da=dimensionArr
      if len(da)==1:return [dv() for _ in range(da[0])]
      else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
   
  def queryInteractive(a, b, c):
      print('? {} {} {}'.format(a, b, c))
      sys.stdout.flush()
      return int(input())
   
  def answerInteractive(ansArr):
      print('! {}'.format(' '.join([str(x) for x in ansArr])))
      sys.stdout.flush()
   
  inf=float('inf')
  # MOD=10**9+7
  # MOD=998244353
  
  from math import gcd,floor,ceil
  import math
  # from math import floor,ceil # for Python2
   
  for _abc in range(1):
      main()