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: def mincoin(arr):
      zero = 0
      num3 = max(arr)//3 - 1
      mid = num3*3+3
      lower = [0,0,0]
      upper = [0,0]
      for i in arr:
          if i < mid:
              lower[i%3] = 1
          elif i == mid:
              zero = 1
          else:
              upper[i%3-1] = 1
      lower[0] = zero
      if sum(upper) == 2:
          return num3 + 3
      elif sum(upper) == 0:
          return min(2,  sum(lower)) + num3
      elif upper[0] == 1:
          if num3 < 0:
              return sum(upper)
          elif min(arr) == 1:
              return num3 + 2 + lower[2]
          else:
              return num3 + 2 + (lower[0] and lower[2] == 1)
      else:
          if lower[1] == 1:
              return num3 + 3
          else:
              return num3 + 2
  
          
  r = int(input())
  res = []
  for i in range(r):
      input()
      arr = list(map(int, input().split()))
      res.append(mincoin(arr))
      
  for i in res:
      print(i)