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 sys
  def ec(n,l1):
      ans1=0
      ans2=0
      ans3=0
      i=0
      while(i<len(l1)):
          if l1[i]%3!=0:
              ans1=sys.maxsize
              break
          i=i+1
      if i==len(l1):
          ans1=max(l1)//3
      i=0
      while(i<len(l1)):
          if l1[i]%3==2:
              ans2=sys.maxsize
              break
          i=i+1
      if i==len(l1):
          ans2=(max(l1)//3)+1
      i=0
      while(i<len(l1)):
          if l1[i]%3==1:
              ans3=sys.maxsize
              break
          i=i+1
      if i==len(l1):
          ans3=(max(l1)//3)+1
      i=0
      ans4=0
      while(i<len(l1)):
          if l1[i]%3==1:
              ans4=max(ans4,l1[i]//3)
          elif l1[i]%3==2:
              ans4=max(ans4,l1[i]//3)
          else:
              ans4=max(ans4,l1[i]//3-1)
          i=i+1
      ans4=ans4+2
      i=0
      ans5=0
      while(i<len(l1)):
          if l1[i]==1:
              ans5=sys.maxsize
              break
          elif l1[i]%3==1:
              ans5=max(ans5,l1[i]//3-1)
          else:
              ans5=max(ans5,l1[i]//3)
          i=i+1
      ans5=ans5+2
      return min(ans1,ans2,ans3,ans4,ans5)
  t=int(input())
  l=[]
  for x in range(t):
      n=int(input())
      s=str(input())
      l.append([s,n])
  la=[]
  for x in l:
      lt=x[0].split(' ',)
      l1=[]
      for y in lt:
          l1.append(int(y))
      n=x[1]
      assert n==len(l1)
      ans=ec(n,l1)
      la.append(ans)
  for x in la:
      print(x)