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;
  #pragma GCC optimize("O3")
  #pragma GCC target("sse4")
  long double pi = 3.141592653589793238462643383279;
  const long long INF = 1e18 + 10;
  template <typename Arg1>
  void __f(const char* name, Arg1&& arg1) {
    std::cerr << name << " : " << arg1 << endl;
  }
  template <typename Arg1, typename... Args>
  void __f(const char* names, Arg1&& arg1, Args&&... args) {
    const char* comma = strchr(names + 1, ',');
    std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
    __f(comma + 1, args...);
  }
  const long long mod = 998244353;
  long long power(long long x, long long y) {
    long long res = 1;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) res = (res * x) % mod;
      y = y >> 1;
      x = (x * x) % mod;
    }
    return res;
  }
  void bruteforce() {}
  void init() {}
  const long long mxN = 1e5 + 10;
  void solve() {
    long long n;
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    bool one = 0;
    long long x = 0, y = 0, z = 0;
    for (long long i = 0; i < n; i++) {
      if (a[i] == 1) one = 1;
      if (a[i] % 3 == 0) {
        x = max(x, a[i]);
      } else if (a[i] % 3 == 1) {
        y = max(y, a[i]);
      } else
        z = max(z, a[i]);
    }
    long long val1 = max({x / 3 - ((y && z) ? 1ll : 0ll), y / 3, z / 3}) +
                     (z ? 1ll : 0ll) + (y ? 1ll : 0ll);
    long long val2 = max({x / 3, y / 3, z / 3}) + 2;
    long long val3 = max({x / 3, (y / 3 - (!one ? 1ll : 0ll)), z / 3}) + 2;
    cout << min({val1, val2, val3}) << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    init();
    long long T = 1;
    cin >> T;
    for (long long i = 1; i <= T; i++) {
      solve();
    }
  }
  