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.*;
  
  public class CF1620D extends PrintWriter {
  	CF1620D() { super(System.out); }
  	Scanner sc = new Scanner(System.in);
  	public static void main(String[] $) {
  		CF1620D o = new CF1620D(); o.main(); o.flush();
  	}
  
  	static final int INF = 2000000000;
  	int solve(int a, int m1, int m2) {
  		int m3 = INF;
  		for (int n1 = 0; n1 <= m1; n1++)
  			for (int n2 = 0; n2 <= m2; n2++) {
  				int b = n1 * 1 + n2 * 2;
  				if (a >= b) {
  					int c = a - b;
  					if (c % 3 == 0)
  						m3 = Math.min(m3, c / 3);
  				}
  			}
  		return m3;
  	}
  	void main() {
  		int t = sc.nextInt();
  		while (t-- > 0) {
  			int n = sc.nextInt();
  			int[] aa = new int[n];
  			for (int i = 0; i < n; i++)
  				aa[i] = sc.nextInt();
  			int m_ = INF;
  			for (int m1 = 0; m1 < 2; m1++)
  				for (int m2 = 0; m2 < 3; m2++) {
  					int m3 = 0;
  					for (int i = 0; i < n; i++)
  						m3 = Math.max(m3, solve(aa[i], m1, m2));
  					m_ = Math.min(m_, m1 + m2 + m3);
  				}
  			println(m_);
  		}
  	}
  }
  