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.math.BigInteger;
  import java.util.*;
  
  public class Main {
      static int MOD = 1000000007;
  
      // After writing solution, quick scan for:
      //   array out of bounds
      //   special cases e.g. n=1?
      //   npe, particularly in maps
      //
      // Big numbers arithmetic bugs:
      //   int overflow
      //   sorting, or taking max, after MOD
      void solve() throws IOException {
          int T = ri();
          for (int Ti = 0; Ti < T; Ti++) {
              int n = ri();
              int[] a = ril(n);
  
              // 3s are good, obviously, can get any mod 3 = 0
              // we at most have a very small number of 1s and 2s, which can replace a 3.
              // i'm conjecturing we need at most 2 1's and 2 2's
  
              int ans = Integer.MAX_VALUE;
              for (int numOnes = 0; numOnes <= 2; numOnes++) {
                  for (int numTwos = 0; numTwos <= 2; numTwos++) {
                      boolean bad = false;
                      int curr = 0;  // number of 3s needed
                      for (int ai : a) {
                          int remainder = ai % 3;
                          if (remainder == 0) {
                              int need = ai;
                              int x = numOnes;
                              int y = numTwos;
                              while (need > 0 && x > 0 && y > 0) {
                                  need -= 3;
                                  x--;
                                  y--;
                              }
                              curr = Math.max(curr, need / 3);
                          } else if (remainder == 1) {
                              int bingo = Integer.MAX_VALUE;
                              // we can either use {1}, {2, 2}
                              if (numOnes > 0) {
                                  int need = ai;
                                  int x = numOnes;
                                  int y = numTwos;
                                  x--;
                                  need -= 1;
                                  while (need > 0 && x > 0 && y > 0) {
                                      need -= 3;
                                      x--;
                                      y--;
                                  }
                                  bingo = Math.min(bingo, need / 3);
                              }
                              if (numTwos >= 2 && ai - 4 >= 0) {
                                  int need = ai;
                                  int x = numOnes;
                                  int y = numTwos;
                                  y -= 2;
                                  need -= 4;
                                  while (need > 0 && x > 0 && y > 0) {
                                      need -= 3;
                                      x--;
                                      y--;
                                  }
                                  bingo = Math.min(bingo, need / 3);
                              }
                              if (bingo == Integer.MAX_VALUE) {
                                  bad = true;
                                  break;
                              } else {
                                  curr = Math.max(curr, bingo);
                              }
                          } else if (remainder == 2) {
                              // we can do {1, 1} or {2}
                              int bingo = Integer.MAX_VALUE;
                              if (numOnes >= 2) {
                                  int need = ai;
                                  int x = numOnes;
                                  int y = numTwos;
                                  x -= 2;
                                  need -= 2;
                                  while (need > 0 && x > 0 && y > 0) {
                                      need -= 3;
                                      x--;
                                      y--;
                                  }
                                  bingo = Math.min(bingo, need / 3);
                              }
                              if (numTwos >= 1) {
                                  int need = ai;
                                  int x = numOnes;
                                  int y = numTwos;
                                  y--;
                                  need -= 2;
                                  while (need > 0 && x > 0 && y > 0) {
                                      need -= 3;
                                      x--;
                                      y--;
                                  }
                                  bingo = Math.min(bingo, need / 3);
                              }
                              if (bingo == Integer.MAX_VALUE) {
                                  bad = true;
                                  break;
                              } else {
                                  curr = Math.max(curr, bingo);
                              }
                          }
                      }
                      if (bad) continue;
                      ans = Math.min(ans, curr + numOnes + numTwos);
                  }
              }
              pw.println(ans);
          }
      }
      // IMPORTANT
      // DID YOU CHECK THE COMMON MISTAKES ABOVE?
  
      // Template code below
  
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter pw = new PrintWriter(System.out);
  
      public static void main(String[] args) throws IOException {
          Main m = new Main();
          m.solve();
          m.close();
      }
  
      void close() throws IOException {
          pw.flush();
          pw.close();
          br.close();
      }
  
      int ri() throws IOException {
          return Integer.parseInt(br.readLine().trim());
      }
  
      long rl() throws IOException {
          return Long.parseLong(br.readLine().trim());
      }
  
      int[] ril(int n) throws IOException {
          int[] nums = new int[n];
          int c = 0;
          for (int i = 0; i < n; i++) {
              int sign = 1;
              c = br.read();
              int x = 0;
              if (c == '-') {
                  sign = -1;
                  c = br.read();
              }
              while (c >= '0' && c <= '9') {
                  x = x * 10 + c - '0';
                  c = br.read();
              }
              nums[i] = x * sign;
          }
          while (c != '\n' && c != -1) c = br.read();
          return nums;
      }
  
      long[] rll(int n) throws IOException {
          long[] nums = new long[n];
          int c = 0;
          for (int i = 0; i < n; i++) {
              int sign = 1;
              c = br.read();
              long x = 0;
              if (c == '-') {
                  sign = -1;
                  c = br.read();
              }
              while (c >= '0' && c <= '9') {
                  x = x * 10 + c - '0';
                  c = br.read();
              }
              nums[i] = x * sign;
          }
          while (c != '\n' && c != -1) c = br.read();
          return nums;
      }
  
      int[] rkil() throws IOException {
          int sign = 1;
          int c = br.read();
          int x = 0;
          if (c == '-') {
              sign = -1;
              c = br.read();
          }
          while (c >= '0' && c <= '9') {
              x = x * 10 + c - '0';
              c = br.read();
          }
          return ril(x);
      }
  
      long[] rkll() throws IOException {
          int sign = 1;
          int c = br.read();
          int x = 0;
          if (c == '-') {
              sign = -1;
              c = br.read();
          }
          while (c >= '0' && c <= '9') {
              x = x * 10 + c - '0';
              c = br.read();
          }
          return rll(x);
      }
  
      char[] rs() throws IOException {
          return br.readLine().toCharArray();
      }
  
      void sort(int[] A) {
          Random r = new Random();
          for (int i = A.length-1; i > 0; i--) {
              int j = r.nextInt(i+1);
              int temp = A[i];
              A[i] = A[j];
              A[j] = temp;
          }
          Arrays.sort(A);
      }
  
      void printDouble(double d) {
          pw.printf("%.16f", d);
      }
  }