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;
  void Err() {}
  template <class T, class... A>
  void Err(T a, A... x) {
    cerr << a << ',';
    Err(x...);
  }
  template <class X, class Y, class... A>
  void Err(pair<X, Y> a, A... x) {
    cerr << '(' << a.first << ',' << a.second << "),";
    Err(x...);
  }
  template <template <class...> class T, class t, class... A>
  void Err(T<t> a, A... x) {
    cerr << a.size() << ":{";
    for (auto v : a) Err(v);
    cerr << "},";
    Err(x...);
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long rnd(long long l, long long r) {
    return uniform_int_distribution<long long>(l, r)(rng);
  }
  template <class T>
  void Min(T &a, const T b) {
    if (a > b) a = b;
  }
  template <class T>
  void Max(T &a, const T b) {
    if (a < b) a = b;
  }
  const long long linf = 0x3f3f3f3f3f3f3f3f;
  const int inf = 0x3f3f3f3f;
  const int N = 2e7 + 10, M = 500;
  const long long mod = 1e9 + 7;
  const long double eps = 1e-8;
  int sgn(long double a) { return a < -eps ? -1 : (a < eps ? 0 : 1); }
  int solve() {
    int n, ret;
    cin >> n;
    vector<int> ar(n);
    for (auto &i : ar) cin >> i;
    sort(ar.begin(), ar.end());
    if (ar.back() % 3 == 2) {
      ret = ar.back() / 3 + 1;
      for (auto i : ar)
        if (i % 3 == 1) {
          ret++;
          break;
        }
    } else if (ar.back() % 3 == 1) {
      ret = ar.back() / 3 + 1;
      bool f1 = 1, f2 = 1;
      for (auto i : ar) {
        if (i % 3 == 2) f1 = 0;
        if (i == ar.back() - 1 || i == 1) f2 = 0;
      }
      if (!f1 && !f2) ret++;
    } else {
      ret = ar.back() / 3;
      for (auto i : ar) {
        if (i % 3) {
          ret++;
          break;
        }
      }
    }
    return ret;
  }
  int main() {
    ;
    ios::sync_with_stdio(0), cin.tie(0);
    int _;
    cin >> _;
    while (_--) {
      cout << solve() << '\n';
    }
  }
  