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,os,io
  input = sys.stdin.readline # for strings
  # input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline # for non-strings
  
  PI = 3.141592653589793238460
  INF =  float('inf')
  MOD  = 1000000007
  # MOD = 998244353
  
  
  def bin32(num):
      return '{0:032b}'.format(num)
  
  def add(x,y):
      return (x+y)%MOD
  
  def sub(x,y):
      return (x-y+MOD)%MOD
  
  def mul(x,y):
      return (x*y)%MOD
  
  def gcd(x,y):
      if y == 0:
          return x
      return gcd(y,x%y)
  
  def lcm(x,y):
      return (x*y)//gcd(x,y)
  
  def power(x,y):
      res = 1
      x%=MOD
      while y!=0:
          if y&1 :
              res = mul(res,x)
          y>>=1
          x = mul(x,x)
          
      return res
          
  def mod_inv(n):
      return power(n,MOD-2)
  
  def prob(p,q):
      return mul(p,power(q,MOD-2))    
    
  def ii():
      return int(input())
  
  def li():
      return [int(i) for i in input().split()]
  
  def ls():
      return [i for i in input().split()]
  
  for t in range(ii()):
      t+=1
      n = ii()
      a = li()
      a = list(set(a))    
      a.sort()
      n = len(a)
      flag = 1
      # if t == 74:
      #     print(a)
      cnt = [0 , 0 , 0 ]
      for i in a:
          if i % 3 == 2 and i % 2 != 0:
              flag = 0
          cnt[i%3]+=1
      
      if cnt[0] == n:
          print(a[-1]//3 )
          continue
      if a[-1]%3 == 0:
          print(a[-1]//3 + 1)
          continue
  
      if cnt[1] == 0 or cnt[2] == 0:
          print(a[-1]//3 + 1)
          continue
  
  
  
      if a[-1] % 3 == 1:
          if 1 in a or a[-1] - 1 in a:
              print(a[-1]//3 + 2)       
          else:
              print(a[-1]//3 + 1)
      else:
          print(a[-1]//3 + 2)
  