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 java.util.*;
  
  public class Main {
      public static void main(String[] args) throws Exception {
          final Scanner cin = new Scanner(System.in);
          final StringBuilder cout = new StringBuilder();
  
          int TC = cin.nextInt();
          for (; TC-- > 0; ) {
              int n = cin.nextInt();
              List<Integer> arr = new ArrayList<>(n);
              for (int i = 0; i < n; i++)
                  arr.add(cin.nextInt());
              int ans = Integer.MAX_VALUE, tans = 0, three = Collections.max(arr) / 3;
              for (int one = 0; one <= 1; one++) { // 1 can occur max once
                  for (int two = 0; two <= 2; two++) {// 2 can occur max twice
                      for (int tc = three; tc >= 0 && tc >= three - 3; tc--) {
                          for (int no : arr) { // try to see if everything can be formed w/ above coins fixed
                              tans = Integer.MAX_VALUE;
                              if (no % 3 == 0) {
                                  if (tc >= no / 3)
                                      tans = one + two + tc;
                                  if (one > 0 && two > 0 && (1 + tc) >= no / 3)
                                      tans = one + two + tc;
                              } else if (no % 3 == 1) {
                                  if (tc >= no / 3 && one > 0)
                                      tans = one + two + tc;
                                  if (tc >= (no - 4) / 3 && two == 2 && no != 1)
                                      tans = one + two + tc;
                              } else if (no % 3 == 2) {
                                  if (tc >= no / 3 && two > 0)
                                      tans = one + two + tc;
                                  if (one > 0 && two == 2 && (1 + tc) >= no / 3)
                                      tans = one + two + tc;
                              }
                              if (tans == Integer.MAX_VALUE)
                                  break;
                          }
                          ans = Math.min(tans, ans);
                      }
                  }
              }
              cout.append(ans).append('\n');
          }
          System.out.println(cout);
      }
  
  }
  