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
  import math
  from collections import deque,Counter
  from sys import stdin
  
  #sys.setrecursionlimit(10**7)
  
  int1=lambda x: int(x)-1
  stinput=lambda :stdin.readline()[:-1]
  ii=lambda :int(stinput())
  mi=lambda :map(int, stdin.readline().split())
  li=lambda :list(mi())
  mi1=lambda :map(int1, stdin.readline().split())
  li1=lambda :list(mi1())
  mis=lambda :map(str, stdin.readline().split())
  lis=lambda :list(mis())
  
  from collections import defaultdict
  #d=defaultdict(int)
  #d=defaultdict(lambda:1)
  
  import heapq
  #heapq.heapify(hq)
  #heapq.heappop(hq)
  #heapq.heappush(hq,1)
  
  mod=10**9+7
  Mod=998244353
  INF=10**18
  ans=0
  
  def f(s):
    cnt=0
    for i in a:
      tmp=INF
      for j in s:
        if i-j>=0 and (i-j)%3==0:
          tmp=min(tmp,(i-j)//3)
      cnt=max(tmp,cnt)
    return cnt    
  
  t=ii()
  for _ in range(t):
    n=ii()
    a=li()
    ans=INF
    for i in range(1,4):
      for j in range(1,4):
        s=set()
        s.add(0)
        for k in range(i):
          for l in range(j):
            s.add(k+l*2)
        s=list(s)
        ans=min(ans,f(s)+i+j-2)
        #print(i,j,s,f(s))
    print(ans)