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 ExactChange
  {
      public static int numcoins(int n)
      {
          int coins = 0;
          if(n%3==0)
              coins = n/3;
          else if(n%3==1)
              coins = (n-1)/3 + 1;
          else
              coins = (n-2)/3 + 1;
          return coins;
      }
  
      public static int numcoinmulti(int a[])
      {
          int len = a.length;
          int m = max_element(a);
          int num = numcoins(m);
          int mod3[] = {0, 0, 0};
          for(int i = 0; i<len; i++)
          {
              mod3[a[i]%3]++;
          }
          if(mod3[m%3]==len)
          {
              return num;  
          }
          else
          {
              if(m%3 == 0)
              {
                  if(mod3[1]!=0 && mod3[2]==0)
                      num++;
                  else if(mod3[2]!=0 && mod3[1]==0)
                      num++;
                  else if(mod3[1]!=0 && mod3[2]!=0)
                      num++;
              }
              else if(m%3 == 1)
              {
                  if(mod3[2]!=0 && mod3[1]==0);
                  else if(mod3[2]!=0 && (freq(a,1)!=0 || freq(a,m-1)!=0))
                  {
                      num++;
                      
                  }
              }
              else
              {
                  if(mod3[1]!=0)
                      num++;
              }
              return num;
          }
      }
  
      public static void main(String[] args)
      {
          Scanner sc = new Scanner(System.in);
          int t = sc.nextInt();
          int n[] = new int[t];
          int ans[] = new int[t];
          for(int i = 0; i<t; i++)
          {
              n[i] = sc.nextInt();
              int a[] = new int[n[i]];
              for(int j = 0; j<n[i]; j++)
              {
                  a[j] = sc.nextInt();
              }
              ans[i] = numcoinmulti(a);
          }
          for(int i = 0; i<t; i++)
          {
              System.out.println(ans[i]);
          }
      }
  
      public static int max_element(int a[])
      {
          int max = a[0];
          for(int i = 0; i<a.length; i++)
          {
              if(a[i]>max)
                  max = a[i];
          }
          return max;
      }
  
      public static int min_element(int a[])
      {
          int min = a[0];
          for(int i = 0; i<a.length; i++)
          {
              if(a[i]<min)
                  min = a[i];
          }
          return min;
      }
      public static int freq(int a[], int x)
      {
          int f = 0;
          for(int i = 0; i<a.length; i++)
          {
              if(a[i]==x)
                  f++;
          }
          return f;
      }
  }
  