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
  tokens = ''.join(sys.stdin.readlines()).split()[::-1]
  
  def next(): return tokens.pop()
  def nextInt(): return int(next())
  def nextFloat(): return float(next())
  def getIntArray(n): return [nextInt() for _ in range(n)]
  def getFloatArray(n): return [nextFloat() for _ in range(n)]
  def getStringArray(n): return [next() for _ in range(n)]
  
  def isPossible(s, ones, twos, threes):
      if s == 0: return True
      elif ones * 1 + twos * 2 + threes * 3 < s: return False
      elif s <= 15: return recurse(s, 0, [ones, twos, min(threes, 5)])
  
      req = (s - 10) // 3
      if threes < req: return False
      threes -= req; s -= req * 3
      return isPossible(s, ones, twos, threes)
  
  def recurse(s, idx, counts):
      if s == 0: return True
      elif idx == len(counts): return False
  
      for i in range(counts[idx] + 1):
          counts[idx] -= i
          if recurse(s - i * (idx + 1), idx + 1, counts):
              return True
          counts[idx] += i
      return False
  
  testcase = True
  def solve(testcase = 1):
      N = nextInt()
      A = getIntArray(N)
      threes = max(i // 3 for i in A)
      ans = 999999999999999999999999999999999999999999999999999999
      for dth in range(4):
          for dtw in range(4):
              for do in range(4):
                  if all(isPossible(ele, do, dtw, threes - dth) for ele in A):
                      ans = min(ans, do + dtw + threes - dth)
      print(ans)
  
  
  if testcase is None:
      testcaseCount = 1
      while tokens:
          solve(testcaseCount)
          testcaseCount += 1
  else:
      testcaseCount = nextInt() if testcase else 1
      for tc in range(testcaseCount):
          solve(tc + 1)
      assert not tokens