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: from sys import stdin, stdout, setrecursionlimit
  from math import gcd, sqrt, factorial, pi, inf
  from collections import deque, defaultdict
  from bisect import bisect, bisect_left
  from time import time
  from itertools import permutations as per
  from heapq import heapify, heappush, heappop, heappushpop
  
  input = stdin.readline
  R = lambda: map(int, input().split())
  I = lambda: int(input())
  S = lambda: input().rstrip('\r\n')
  L = lambda: list(R())
  P = lambda x: stdout.write(str(x) + '\n')
  lcm = lambda x, y: (x * y) // gcd(x, y)
  nCr = lambda x, y: (f[x] * inv((f[y] * f[x - y]) % N)) % N
  inv = lambda x: pow(x, N - 2, N)
  sumx = lambda x: (x ** 2 + x) // 2
  N = 10 ** 9 + 7
  
  def ok(x):
      for i1 in range(i+1):
          for j1 in range(j+1):
              if i1+j1*2>x:continue
              if (x-i1-j1*2)%3:continue
              if (x-i1-j1*2)//3<=k:
                  return True
      return False
  
  for _ in range(I()):
      n=I()
      a=sorted(set(R()))
      f=1
      ans=inf
      m=max(a)
      for i in range(4):
          for j in range(4):
              k=max(0,(m+2-i-j*2)//3)
              if k*3+j*2+i<m:continue
              f=1
              for el in a:
                  if not ok(el):
                      f=0
                      break
              if f:
                  ans=min(ans,k+j+i)
                  #print(ans,k,j,i)
      print(ans)