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: #include <bits/stdc++.h>
  using namespace std;
  const int MOD = 1000000007;
  void solve() {
    int n;
    cin >> n;
    vector<int> arr(n);
    for (int i = 0; i < n; ++i) cin >> arr[i];
    sort(arr.begin(), arr.end());
    set<int> s;
    for (int num : arr) s.insert(num);
    if (n == 1)
      cout << arr[0] / 3 + (arr[0] % 3 == 0 ? 0 : 1) << endl;
    else {
      if (arr[0] == 1) {
        if (arr[1] == 2) {
          if (arr[n - 1] % 3 == 0)
            cout << arr[n - 1] / 3 + 1 << endl;
          else
            cout << arr[n - 1] / 3 + 2 << endl;
        } else {
          bool two = false;
          for (int num : arr)
            if (num % 3 == 2) {
              two = true;
              break;
            }
          if (two && arr[n - 1] % 3 == 0)
            cout << arr[n - 1] / 3 + 1 << endl;
          else if (two)
            cout << arr[n - 1] / 3 + 2 << endl;
          else
            cout << arr[n - 1] / 3 + 1 << endl;
        }
      } else {
        set<int> mods;
        for (int num : arr)
          if (num % 3) mods.insert(num % 3);
        if (mods.size() == 0)
          cout << arr[n - 1] / 3 << endl;
        else if (mods.size() == 1)
          cout << arr[n - 1] / 3 + 1 << endl;
        else if ((!s.count(arr[n - 1] / 3 * 3) && arr[n - 1] % 3 == 1) ||
                 arr[n - 1] % 3 == 0)
          cout << arr[n - 1] / 3 + 1 << endl;
        else
          cout << arr[n - 1] / 3 + 2 << endl;
      }
    }
  }
  int main() {
    int t;
    cin >> t;
    for (int i = 0; i < t; ++i) solve();
    return 0;
  }
  