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 MyClass {
      public static void main(String args[]) {
          /*
          Easy idea but
          One stupid mistake: exist number 1 => must use 1 coin
          */
        Scanner in= new Scanner(System.in);
        int t= in.nextInt();
        while(t-->0){
            int n= in.nextInt();
              int[] a= new int[n];
              for(int i=0;i<n;i++) a[i]= in.nextInt();
              int[] div= new int[3];
              boolean existOne= false;
              for(int num: a){
                  div[num%3]= Math.max(div[num%3], num);
                  if(num==1) existOne= true;
              }
              int res=Integer.MAX_VALUE;
              // div[1]= 1+3
              int three= Math.max(div[0]/3,Math.max(div[1]/3,div[2]/3));
              int one=0, two=0;
              if(div[2]>0) two=1;
              if(div[1]>0) one=1;
              if(one>0 && two>0 && div[0]>0) three= Math.max(Math.max(0,div[0]/3-1),Math.max(div[1]/3,div[2]/3));
              int res1= one+two+three;
              one=0;two=0;three=0;
              //div[1]= 2+2
              
              int res2= Integer.MAX_VALUE;
              if(div[1]>=4){
                  three=  Math.max(div[0]/3,Math.max(Math.max(0,div[1]-4)/3,div[2]/3));
                  two=2;
                  res2= one+two+three;
                  if(existOne) res2++;
              }
  
              System.out.println(Math.min(res1,res2));
        }
      }
  }