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;
  const int N = 1e5 + 5;
  const int mod = 1e9 + 7;
  int T, n;
  vector<int> v;
  map<int, bool> m;
  bool check(int a, int b, int c) {
    for (int i = 0; i < v.size(); i++) {
      int val = v[i];
      m.clear();
      for (int A = 0; A <= a; A++) {
        for (int B = 0; B <= b; B++) {
          for (int C = max(val / 3 - 2, 0); C <= min(val / 3, c); C++) {
            m[A + B * 2 + C * 3] = true;
          }
        }
      }
      if (!m[val]) return false;
    }
    return true;
  }
  void solve() {
    cin >> n;
    v.clear();
    m.clear();
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      if (!m[x]) v.push_back(x);
      m[x] = true;
    }
    int ans = 1e9;
    n = v.size();
    sort(v.begin(), v.end());
    for (int a = 0; a <= 2; a++) {
      for (int b = 0; b <= 2; b++) {
        for (int c = max(v.back() / 3 - 2, 0); c <= v.back() / 3; c++) {
          if (check(a, b, c)) {
            ans = min(ans, a + b + c);
          }
        }
      }
    }
    cout << ans << "\n";
    return;
  }
  signed main() {
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    cin >> T;
    while (T--) solve();
    return 0;
  }
  