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;
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int tc;
    cin >> tc;
    for (int i = 1; i <= tc; i++) {
      int n;
      cin >> n;
      vector<int> a(n);
      int coins = 1e9 + 7;
      for (int i = 0; i < n; i++) {
        cin >> a[i];
      }
      auto check = [&](int one, int two, int val) -> int {
        vector<int> combination;
        for (int i = 0; i <= one; i++) {
          for (int j = 0; j <= two; j++) {
            combination.emplace_back(i + 2 * j);
          }
        }
        int cur_coins = 1e9 + 7;
        for (int i : combination) {
          if ((val < i) or (val - i) % 3 != 0) continue;
          cur_coins = min(cur_coins, (val - i) / 3);
        }
        return cur_coins + one + two;
      };
      for (int one = 0; one < 3; one++) {
        for (int two = 0; two < 3; two++) {
          int cur = 0;
          for (int i : a) {
            cur = max(cur, check(one, two, i));
          }
          coins = min(coins, cur);
        }
      }
      cout << coins << '\n';
    }
    return 0;
  }
  