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: import io,os
  input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
  
  def main(t):
  
  
      n = int(input())
      arr = list(map(int,input().split()))
  
      maximum = max(arr)
      ans = maximum//3 + (maximum%3!=0)
  
  
      if maximum%3==0:
          for i in range(n):   
              if arr[i]%3!=0:  
                  ans += 1
                  break
  
      elif maximum%3==2:
          for i in range(n):   
              if arr[i]%3==1:  
                  ans += 1
                  break 
  
      else:
          op = [1,1]
  
          for i in range(n):
              if arr[i]==maximum:  continue 
  
              if arr[i]%3==2:  op[0] = 0
              if arr[i]==1 or arr[i]==maximum-1:  op[1] = 0
  
          if sum(op)==0:  ans += 1
  
          
                 
  
  
          
  
  
  
  
      print(ans)
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  T = int(input())
  t = 1
  while t<=T:
      main(t)
      t += 1
  