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>
  #pragma GCC optimize("O3")
  #pragma GCC target("sse4")
  using namespace std;
  void __print(long long x) { cerr << x; }
  void __print(int32_t x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const char *x) { cerr << '\"' << x << '\"'; }
  void __print(const string &x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cerr << '{';
    __print(x.first);
    cerr << ',';
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T &x) {
    long long f = 0;
    cerr << '{';
    for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  const long long MOD = 1e9 + 7;
  const long long MOD1 = 998244353;
  const long long N = 2e5 + 5;
  const long long INF = 1000111000111000111LL;
  const long double EPS = 1e-12;
  const long double PI = 3.141592653589793116;
  bool can(long long c1, long long c2, long long c3, long long n) {
    long long val = n / 3;
    if (val >= 0 && val <= c3) {
      long long num = n - val * 3;
      for (long long i = 0; i <= c1; i++) {
        for (long long j = 0; j <= c2; j++) {
          if (i + 2 * j == num) {
            return true;
          }
        }
      }
    }
    val--;
    if (val >= 0 && val <= c3) {
      long long num = n - val * 3;
      for (long long i = 0; i <= c1; i++) {
        for (long long j = 0; j <= c2; j++) {
          if (i + 2 * j == num) {
            return true;
          }
        }
      }
    }
    val--;
    if (val >= 0 && val <= c3) {
      long long num = n - val * 3;
      for (long long i = 0; i <= c1; i++) {
        for (long long j = 0; j <= c2; j++) {
          if (i + 2 * j == num) {
            return true;
          }
        }
      }
    }
    return false;
  }
  void solve_test() {
    long long ans = INF;
    long long n;
    cin >> n;
    long long mx = 0;
    long long a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      mx = max(a[i], mx);
    }
    for (long long c3 = mx / 3 + 1; c3 >= max(0LL, mx / 3 - 2); c3--) {
      for (long long c2 = 0; c2 <= 3; c2++) {
        for (long long c1 = 0; c1 <= 3; c1++) {
          bool ok = true;
          for (long long i = 0; i < n; i++) {
            ok &= can(c1, c2, c3, a[i]);
          }
          if (ok) {
            ans = min(ans, c1 + c2 + c3);
          }
        }
      }
    }
    cout << ans << '\n';
  }
  int32_t main() {
    std::ios_base::sync_with_stdio(0);
    cout.tie(nullptr);
    long long T;
    cin >> T;
    for (long long tc = 1; tc <= T; tc++) {
      solve_test();
    }
    return 0;
  }
  