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 target("avx")
  #pragma GCC optimize("O3")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  using ll = long long;
  using datas = pair<ll, ll>;
  using ddatas = pair<long double, long double>;
  using tdata = pair<ll, datas>;
  using vec = vector<ll>;
  using mat = vector<vec>;
  using pvec = vector<datas>;
  using pmat = vector<pvec>;
  constexpr ll mod = 1000000007;
  constexpr ll inf = 1LL << 60;
  constexpr long double eps = 1e-9;
  const long double PI = acosl(-1);
  template <class T, class E>
  ostream& operator<<(ostream& os, const pair<T, E>& p) {
    return os << "(" << p.first << "," << p.second << ")";
  }
  template <class T>
  ostream& operator<<(ostream& os, const vector<T>& v) {
    os << "{";
    bool f = false;
    for (auto& x : v) {
      if (f) os << ",";
      os << x;
      f = true;
    }
    os << "}";
    return os;
  }
  template <class T>
  ostream& operator<<(ostream& os, const set<T>& v) {
    os << "{";
    bool f = false;
    for (auto& x : v) {
      if (f) os << ",";
      os << x;
      f = true;
    }
    os << "}";
    return os;
  }
  template <class T>
  ostream& operator<<(ostream& os, const multiset<T>& v) {
    os << "{";
    bool f = false;
    for (auto& x : v) {
      if (f) os << ",";
      os << x;
      f = true;
    }
    os << "}";
    return os;
  }
  template <class T, class E>
  ostream& operator<<(ostream& os, const map<T, E>& v) {
    os << "{";
    bool f = false;
    for (auto& x : v) {
      if (f) os << ",";
      os << x;
      f = true;
    }
    os << "}";
    return os;
  }
  template <class T>
  inline bool chmax(T& a, const T b) {
    bool x = a < b;
    if (x) a = b;
    return x;
  }
  template <class T>
  inline bool chmin(T& a, const T b) {
    bool x = a > b;
    if (x) a = b;
    return x;
  }
  inline void startupcpp(void) noexcept {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
  }
  ll modinv(ll a, const ll m = mod) noexcept {
    ll b = m, u = 1, v = 0, t;
    while (b) {
      t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    return (u + m) % m;
  }
  ll moddevide(const ll a, const ll b) noexcept { return (a * modinv(b)) % mod; }
  vec modncrlistp, modncrlistm;
  ll modncr(const ll n, const ll r) noexcept {
    if (n < r || r < 0) return 0;
    ll i, size = modncrlistp.size();
    if (size <= n) {
      modncrlistp.resize(n + 1);
      modncrlistm.resize(n + 1);
      if (!size) {
        modncrlistp[0] = modncrlistm[0] = 1;
        size++;
      }
      for (i = size; i < (ll)n + 1; ++i)
        modncrlistp[i] = modncrlistp[i - 1] * i % mod;
      modncrlistm[n] = modinv(modncrlistp[n]);
      for (i = n; i > size; --i) modncrlistm[i - 1] = modncrlistm[i] * i % mod;
    }
    return modncrlistp[n] * modncrlistm[r] % mod * modncrlistm[n - r] % mod;
  }
  ll modpow(ll a, ll n, const ll m = mod) {
    if (n < 0) return 0;
    ll res = 1;
    while (n > 0) {
      if (n & 1) res = res * a % m;
      a = a * a % m;
      n >>= 1;
    }
    return res;
  }
  constexpr ll gcd(const ll a, const ll b) noexcept {
    return (!b) ? abs(a) : (a % b == 0) ? abs(b) : gcd(b, a % b);
  }
  constexpr ll lcm(const ll a, const ll b) noexcept { return a / gcd(a, b) * b; }
  ll N, M, K, H, W, A, B, C, D;
  string s, t;
  ll ans;
  int main() {
    startupcpp();
    int codeforces;
    cin >> codeforces;
    while (codeforces--) {
      ll i, j = 0;
      cin >> N;
      vec v(N);
      for (i = 0; i < (ll)N; ++i) {
        cin >> v[i];
        j |= 1 << (v[i] % 3);
      }
      sort((v).begin(), (v).end()),
          (v).erase(unique((v).begin(), (v).end()), (v).end());
      if (__builtin_popcountll(j) == 1)
        cout << (v.back() + 2) / 3 << "\n";
      else if (j == 3 || j == 5)
        cout << v.back() / 3 + 1 << "\n";
      else if (v.back() % 3 != 1 || v[0] == 1 ||
               count((v).begin(), (v).end(), v.back() - 1))
        cout << (v.back() + 2) / 3 + 1 << "\n";
      else
        cout << (v.back() + 2) / 3 << "\n";
    }
  }
  