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 __future__ import division, print_function
  from math import inf
  import itertools
  import sys
  from typing import Counter
  
  from collections import Counter
  
  if sys.version_info[0] < 3:
      input = raw_input
      range = xrange
  
      filter = itertools.ifilter
      map = itertools.imap
      zip = itertools.izip
  
  read_int_vec = lambda x: [int(c) for c in x.split(" ")]
  
  from math import inf, sqrt
  from itertools import groupby
  
  def possible(val,c1,c2,c3):
      for ones in range(0,c1+1):
          for twos in range(0, c2 + 1):
              if ones + 2 * twos > val:
                  continue
              if (val - ones - 2 * twos) % 3 != 0:
                  continue
              if (val - ones - 2 * twos) // 3 <= c3:
                  return True
      return False
  
  def solution(A):
      m = max(A)
      MAXC = 3 
      ans = inf
      for c1 in range(MAXC):
          for c2 in range(MAXC):
              c3 = max(0, (m - c1 - 2 * c2 + 2) // 3)
              assert c1 + 2 * c2 + 3 * c3 >= m
              if all(possible(a,c1,c2,c3) for a in A):
                  ans = min(ans, c1 + c2 + c3)
      return ans
  
  if __name__ == "__main__":
      """
      test(2,1,7,4, 8, 5, 4, 2)
      test(4,5,7,8, 8, 10, 8, 5)
      test(2,2,5,4, 5, 4, 3, 3)
      test(0,3,1,6, 1, 8, 1, 5)
      """
      n_tests = int(input())
  
      for _ in range(n_tests):
          n = int(input())
          A  = read_int_vec(input())
          print(solution(A))