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 _print(long long int t) { cerr << t; }
  void _print(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T>
  void _print(unordered_set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T, class V>
  void _print(multimap<T, V> v);
  template <class T, class V>
  void _print(unordered_map<T, V> v);
  template <class T, class V>
  void _print(unordered_multimap<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T>
  void _print(unordered_multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(unordered_set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(unordered_multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(multimap<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(unordered_map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(unordered_multimap<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <typename T>
  T gcd(T a, T b) {
    if (a % b) return gcd(b, a % b);
    return b;
  }
  template <typename T>
  T lcm(T a, T b) {
    return (a * (b / gcd(a, b)));
  }
  template <typename T>
  void inline in(vector<T> &v, long long int n) {
    v.resize(n);
    for (long long int i = 0; i < n; i++) cin >> v[i];
  }
  bool inline isVowel(char c) {
    return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u');
  }
  long long int n;
  vector<long long int> a;
  bool pp(long long int u, long long int c1, long long int c2, long long int c3) {
    for (int curn1 = (0); curn1 <= (c1); curn1++) {
      for (int curn2 = (0); curn2 <= (c2); curn2++) {
        if (2 * curn2 + curn1 > u) continue;
        if ((u - curn1 - 2 * curn2) % 3 != 0) continue;
        if ((u - curn1 - 2 * curn2) / 3 <= c3) return true;
      }
    }
    return false;
  }
  bool possible(long long int c1, long long int c2, long long int c3) {
    for (auto u : a)
      if (!pp(u, c1, c2, c3)) return false;
    return true;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout << setprecision(8) << fixed;
    long long int txtc;
    cin >> txtc;
    for (int tcase = 1; tcase <= txtc; tcase++) {
      cin >> n;
      in(a, n);
      long long int m = *max_element(a.begin(), a.end());
      ;
      long long int ans = 1e18;
      const long long int maxn = 2;
      for (int coins1 = (0); coins1 <= (3); coins1++) {
        for (int coins2 = (0); coins2 <= (3); coins2++) {
          long long int coins3 = max(0ll, (m - coins1 - 2 * coins2 + 2) / 3);
          if ((3 * coins3 + 2 * coins2 + coins1) < m) continue;
          if (possible(coins1, coins2, coins3)) {
            ans = min(ans, coins2 + coins1 + coins3);
          }
        }
      }
      cout << ans << "\n";
    }
    return 0;
  }
  