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 A, class B>
  ostream& operator<<(ostream& out, const pair<A, B>& p) {
    return out << "(" << p.x << ", " << p.y << ")";
  }
  template <class A>
  ostream& operator<<(ostream& out, const vector<A>& v) {
    for (int i = int(0); i < int(int((v).size())); i++) {
      if (i) out << " ";
      out << v[i];
    }
    return out;
  }
  int n;
  vector<int> a;
  inline bool read() {
    if (!(cin >> n)) return false;
    a.resize(n);
    for (int i = int(0); i < int(n); i++) cin >> a[i];
    return true;
  }
  bool p(int val, int c1, int c2, int c3) {
    for (int cur1 = int(0); cur1 < int(c1 + 1); cur1++)
      for (int cur2 = int(0); cur2 < int(c2 + 1); cur2++) {
        if (cur1 + 2 * cur2 > val) continue;
        if ((val - cur1 - 2 * cur2) % 3 != 0) continue;
        if ((val - cur1 - 2 * cur2) / 3 <= c3) return true;
      }
    return false;
  }
  bool possible(int c1, int c2, int c3) {
    for (int v : a) {
      if (!p(v, c1, c2, c3)) return false;
    }
    return true;
  }
  inline void solve() {
    int m = *max_element(a.begin(), a.end());
    int ans = int(1e9);
    const int MAG = 3;
    for (int c1 = int(0); c1 < int(MAG); c1++)
      for (int c2 = int(0); c2 < int(MAG); c2++) {
        int c3 = max(0, (m - c1 - 2 * c2 + 2) / 3);
        assert(c1 + 2 * c2 + 3 * c3 >= m);
        if (possible(c1, c2, c3)) ans = min(ans, c1 + c2 + c3);
      }
    cout << ans << endl;
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      read();
      solve();
    }
    return 0;
  }
  