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: #!/usr/bin/env python3
  # from typing import *
  
  import sys
  import io
  import math
  import collections
  import decimal
  import itertools
  import bisect
  import heapq
  
  
  def input():
      return sys.stdin.readline()[:-1]
  
  
  # sys.setrecursionlimit(1000000)
  
  # _INPUT = """1
  # 4
  # 2 4 6 7
  # """
  # sys.stdin = io.StringIO(_INPUT)
  
  INF = 10**10
  
  def solve(N, A):
      s = set(A)
      m = max(A)
      T = [False] * 3
      for a in A:
          if a == m:
              continue
          T[a%3] = True
  
      if m == 1:
          return 1
      elif m == 2:
          if T[1]:
              return 2
          else:
              return 1
      
      n = m // 3
      if m % 3 == 0:
          if (not T[1]) and (not T[2]):
              return n
          elif T[1] and (not T[2]):
              return n+1
          elif (not T[1]) and T[2]:
              return n+1
          else:
              return n+1
      elif m % 3 == 1:
          if (not T[1]) and (not T[2]):
              return n+1
          elif T[1] and (not T[2]):
              return n+1
          elif (not T[1]) and T[2]:
              if m-1 in s:
                  return n+2
              else:
                  return n+1
          else:
              if 1 not in s and m-1 not in s:
                  return n+1
              else:
                  return n+2
      else:
          if (not T[1]) and (not T[2]):
              return n+1
          elif T[1] and (not T[2]):
              return n+2
          elif (not T[1]) and T[2]:
              return n+1
          else:
              return n+2
  
  def solve_naive(N, A):
      ans = INF
      for i1, i2, i3 in itertools.product(range(6), repeat=3):
          ok = True
          for a in A:
              ok1 = False
              for u1 in range(i1+1):
                  for u2 in range(i2+1):
                      for u3 in range(i3+1):
                          if a == u1 + u2*2 + u3*3:
                              ok1 = True
                              break
              if not ok1:
                  ok = False
                  break
          if ok:
              ans = min(ans, i1+i2+i3)
      return ans
              
  
  # N = 4
  # for A in itertools.product(range(1, 11), repeat=4):
  #     ans1 = solve(N, A)
  #     ans2 = solve_naive(N, A)
  #     if ans1 != ans2:
  #         print('NG', A, ans1, ans2)
  # sys.exit()
  
  
  
  T0 = int(input())
  for _ in range(T0):
      N = int(input())
      A = list(map(int, input().split()))
      print(solve(N, A))