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: t = int(input())
  
  
  def get_sum_3(count):
      return count[3] + min(count[2], count[1]) + (max(0, count[1] - count[2]) // 3)
  
  for i in range(t):
      count = [0, 0, 0, 0]
      n = int(input())
      a = [int(x) for x in input().split(' ')]
      a.sort()
      a = a[::-1]
      may3 = 0
      for j in range(n):
          div = a[j] // 3
          res = a[j] % 3
          if(res == 0 and may3 == 0):
              may3 = div
  
          if(get_sum_3(count) < div):
              count[3] += div - get_sum_3(count)
          if(res > 0 and count[res] == 0):
              count[res] = 1
  
      if(a[0] % 3 == 0 and count[1] and count[2]):
          count[3] = max(count[3] - 1, 0)
      elif(a[0] % 3 == 1 and count[1] and count[2] and a[-1] != 1 and (may3 == 0 or may3 <= count[3] - 1)):
          count[3] -= 1
          count[1] -= 1
          count[2] += 1
  
      print(count[3] + count[2] + count[1])
  
  