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;
  template <class T>
  ostream &operator<<(ostream &os, const vector<T> &p) {
    os << "[";
    for (auto &it : p) os << it << " ";
    return os << "]";
  }
  template <class S, class T>
  ostream &operator<<(ostream &os, const pair<S, T> &p) {
    return os << "(" << p.first << "," << p.second << ")";
  }
  template <class T>
  inline T Bit(T x, long long i) {
    return (x >> i) & 1;
  }
  template <class T>
  bool umin(T &a, T b) {
    return a > b ? (a = b, true) : false;
  }
  template <class T>
  bool umax(T &a, T b) {
    return a < b ? (a = b, true) : false;
  }
  const long long N = 2e5 + 3;
  void solve(long long tc) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long ans = 1e12;
    for (long long one = 0; one < 3; one++) {
      for (long long two = 0; two < 3; two++) {
        long long cur = 0;
        for (long long i = 0; i < n; i++) {
          long long need = 1e12;
          for (long long j = 0; j <= one; j++) {
            for (long long k = 0; k <= two; k++) {
              if (a[i] - j - k * 2 >= 0 && (a[i] - j - k * 2) % 3 == 0) {
                umin(need, (a[i] - j - k * 2) / 3);
              }
            }
          }
          umax(cur, need);
        }
        umin(ans, cur + one + two);
      }
    }
    {
      cout << ans << '\n';
      return;
    };
  }
  signed main() {
    ios_base ::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    cin >> t;
    for (long long i = 1; i <= t; i++) solve(i);
    return 0;
  }
  