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 static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  import java.util.*;
  import java.io.*;
  
  public class Problem_D {
  	private final static int INF = (int) 1e9 + 5;
  	
  	public static void main(String[] dummy) {
  		MyScanner ms = new MyScanner();
  		PrintStream out = new PrintStream(System.out);
  		
  		int testcases = ms.nextInt();
  		while (testcases-- > 0) {
  			int N = ms.nextInt();
  			int []A = ms.readArray(N);
  			int min_coin = INF;
  			
  			for (int ones = 0; ones <= 1; ones++) {
  				for (int twos = 0; ones + twos <= 2; twos++) {
  					int threes = 0;
  					
  					for (int a : A) {
  						int max_three = INF;
  						
  						for (int i = 0; i <= ones; i++) {
  							for (int j = 0; j <= twos; j++) {
  								int x = a - i - (2 * j);
  								
  								if (x >= 0 && x % 3 == 0) 
  									max_three = min(max_three, x / 3);
  							}
  						}
  						threes = max(threes, max_three);
  					}
  					min_coin = min(min_coin, ones + twos + threes);
  				}
  			}
  			out.println(min_coin);
  		}
  		
  // 		double time = System.currentTimeMillis();
  // 		System.err.println("\nExecution time: " + time + "s");
  		out.flush();
  		out.close();
  	}
  	
  	public static void Sort(int[] arr) { // TODO : Collections.sort always uses Merge sort
  		ArrayList<Integer> sorted = new ArrayList<>();
  		for (int x : arr) 
  			sorted.add(x);
  		
  		Collections.sort(sorted);
  		for (int i = 0; i < arr.length; i++)
  			arr[i] = sorted.get(i);
  	}
  	
  	static class MyScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st = new StringTokenizer("");
  		
  		String next() {
  			while (!st.hasMoreTokens()) 
  				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());
  		}
  		
  		int[] readArray(int n) {
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++) 
  				a[i] = nextInt();
  			return a;
  		}
  	}
  }
  