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: #     __  __                __    _ __      ______                
  #    / / / /___ ___________/ /_  (_) /_    / ____/___ __________ _
  #   / /_/ / __ `/ ___/ ___/ __ \/ / __/   / / __/ __ `/ ___/ __ `/
  #  / __  / /_/ / /  (__  ) / / / / /_    / /_/ / /_/ / /  / /_/ / 
  # /_/ /_/\__,_/_/  /____/_/ /_/_/\__/____\____/\__,_/_/   \__, /  
  #                                  /_____/               /____/   
  
  from collections import Counter
  from sys import *
  from collections import defaultdict as dd
  from math import *
  
  # sys.stdin = open('input.txt', 'r')  
  # sys.stdout = open('output.txt', 'w')
  
  #input functions
  def inp():
      return int(stdin.readline().strip())
  def vinp():
      return map(int,stdin.readline().strip().split())
  def linp():
      return list(map(int,stdin.readline().strip().split()))
  def sinp(n = 1):
      if n==1:
          return stdin.readline().strip()
      elif n==2:
          return list(stdin.readline().strip())
      else:
          return list(stdin.readline().split())
  
  #output function
  def pr(*x, end = "\n"):
      print(*x,end = end)
  
  #others
  def mod(f, val = 1000000007):
      return f % val
  def csort(c):
      sorted(c.items(), key=lambda pair: pair[1], reverse=True)
  def indc(l,n):
      c={}
      for i in range(n):
          c[l[i]]=c.get(l[i],[])+[i+1]
      return c
  
  if __name__ =="__main__":
      cou=inp()
      for _ in range(cou):
          n = inp()
          l = linp()
          m1,m2,m3,m4,m5 = 0,0,0,0,0
          m1t,m2t,m3t,m4t,m5t = True,True,True,True,True
          for i in range(n):
              if (l[i] % 3 != 0):
                  m1 = maxsize
                  m1t = False 
              if l[i] % 3 == 2 :
                  m2 = maxsize
                  m2t = False
              if l[i] % 3 == 1 :
                  m3 = maxsize
                  m3t = False
              if l[i] == 1:
                  m5 = maxsize
                  m5t = False
              if m1t:
                  m1 = max(m1, l[i]//3)
              if m2t:
                  m2 = max(m2, l[i] // 3)
              if m3t:
                  m3 = max(m3, l[i] // 3)
              if l[i] % 3 == 0:
                  m4 = max(m4, l[i] // 3 - 1);
              else:
                  m4 = max(m4, l[i] // 3)
              if m5t:
                  if l[i] % 3 == 1:
                      m5 = max(m5, (l[i] -4) // 3);
                  else:
                      m5 = max(m5, l[i] // 3)
          if m2t:
              m2+=1
          if m3t:
              m3+=1
          if m5t:
              m5+=2
          m4+=2
          pr(min(m1,m2,m3,m4,m5))