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.*;
  import java.util.*;
  import static java.lang.Math.*;
  
  public class D{
  	static int n;
  	static long[] a;
  	static boolean works(long numCoins) {
  		for (int i = 0; i <= 2; i++) {
  			for (int j = 0; j <= 3; j++) {
  				if (numCoins - i - j < 0) continue;
  				if(solve(j, i, numCoins-i-j)) return true;
  			}
  		}
  		return false;
  	}
  	
  	static boolean solve(long... c) {
  		 
  		List<Long> poss = new ArrayList<Long>();
  		for (int i = 0; i <= c[0]; i++) {
  			for (int j = 0; j <= c[1]; j++) {
  				poss.add((long)(i+2*j));
  			}
  		}
  		outer: for (int i = 0; i < a.length; i++) {
  			// Try to make this shit work
  			for (long x: poss) {
  				long temp = a[i]-x;
  				if (temp%3==0) {
  					if (temp >= 0 && temp/3 <= c[2]) continue outer;
  				}
  			}
  			return false;
  		}
  		return true;
  	}
  	
  	public static void main(String[] args) throws IOException {
  		// br = new BufferedReader(new FileReader(".in"));
  		// out = new PrintWriter(new FileWriter(".out"));
  		//new Thread(null, new (), "peepee", 1<<28).start();
  		int t = readInt();
  		while(t-->0) {
  			n = readInt();
  			a = new long[n];
  			// You need to replace at most 2 3s for 3 twos  and 1 three for three ones
  			for (int i =0 ; i < n; i++) a[i] =readInt();
  			
  			long l = 0, r = (int)1e9+10; 
  			while(l+1<r) {
  				long mid = (l+r)/2;
  				if (works(mid)) r = mid;
  				else l =mid;
  			}
  			out.println(r);
  			
  			
  		}
  		
  		
  		out.close();
  	}
  	
  	/* Stupid things to try if stuck:
  	 * n=1, expand bsearch range
  	 * brute force small patterns
  	 * submit stupid intuition
  	 */
  	
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
  	static StringTokenizer st = new StringTokenizer("");
  	static String read() throws IOException{return st.hasMoreTokens() ? st.nextToken():(st = new StringTokenizer(br.readLine())).nextToken();}
  	static int readInt() throws IOException{return Integer.parseInt(read());}
  	static long readLong() throws IOException{return Long.parseLong(read());}
  	static double readDouble() throws IOException{return Double.parseDouble(read());}
  }