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.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  public class D {
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader()
          {
              br = new BufferedReader(
                      new InputStreamReader(System.in));
          }
  
          String next()
          {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() { return Integer.parseInt(next()); }
  
          long nextLong() { return Long.parseLong(next()); }
  
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
  
          String nextLine()
          {
              String str = "";
              try {
                  str = br.readLine();
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
      public static void main(String[] args){
          FastReader sc = new FastReader();
          int t = sc.nextInt();
          StringBuilder s = new StringBuilder();
          while (t-->0){
              int n = sc.nextInt();
              int[] mod3 = new int[n];
              int[] p = new int[n];
              boolean has1 = false;
              boolean has2 = false;
              boolean has_1 =false;
              int maxp = Integer.MIN_VALUE;
              for (int i=0;i<n;i++){
                  p[i] = sc.nextInt();
                  mod3[i] = p[i]%3;
                  if (mod3[i]==1){
                      has1 = true;
                  }
                  if (mod3[i]==2){
                      has2 = true;
                  }
                  if (p[i]==1){
                  has_1 = true;
              }
                  maxp = Math.max(maxp,p[i]);
              }
  
              int mod3_M = maxp%3;
              int min_count = maxp/3;
  
  
              if (mod3_M!=0){
                  min_count ++;
                  if (mod3_M==1&&has2){
                      boolean flag = false;
                      for (int i=0;i<n;i++){
                          if (p[i]==maxp-1){
                              flag = true;
                          }
                      }
                      if (has_1||flag){
                          min_count++;
                      }
                  }else if (mod3_M==2&&has1){
                      min_count++;
                  }
              }
              if (mod3_M==0){
                  if (has1||has2){
                      min_count++;
                  }
              }
              s.append(min_count+"\n");
          }
          System.out.println(s);
      }
  }
  