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: for _ in range(int(input())):
      n = int(input())
      l = list(map(int, input().split()))
      l6 = all([x < 6 for x in l])
      b = 2e9
      for i in range(7):
          for j in range(4):
              for k in range(3):
                  ok = [True, i > 0, i >= 2 or j >= 1,
                        k >= 1 or (i >= 1 and j >= 1) or (i >= 3),
                        (k >= 1 and i >= 1) or (j >= 2) or (j >= 1 and i >= 2) or
                        (i >= 4),
                        (k >= 1 and i >= 2) or (k >= 1 and j >= 1) or (j >= 2 and i >= 1)
                        or (j >= 1 and i >= 3) or (i >= 5), i + 2 * j + 3 * k == 6]
                  if i + 2 * j + 3 * k > 6:
                      continue
                  s = i + j + k
                  if l6:
                      if any([not ok[x] for x in l]):
                          continue
                      b = min(b, s)
                  else:
                      no = False
                      m = 0
                      for x in l:
                          yes = False
                          for I in range(min(6, x), -1, -1):
                              if ok[I] and (x - I) % 3 == 0:
                                  yes = True
                                  m = max(m, (x - I) // 3)
                                  break
                          if not yes:
                              no = True
                              break
                      if no:
                          continue
                      b = min(b, s + m)
      print(b)