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;
  int n;
  vector<int> a;
  void solve() {
    cin >> n;
    a.resize(n);
    for (auto& x : a) {
      cin >> x;
    }
    int cnt = *max_element(a.begin(), a.end()) / 3, ans = INT_MAX;
    for (int i = max(0, cnt - 10); i <= cnt + 10; i++) {
      int twos = 0;
      for (auto x : a) {
        int cur = min(i, x / 3);
        int rem = x - cur - cur - cur;
        twos = max(twos, (rem / 2));
      }
      for (int j = max(0, twos - 10); j <= twos + 10; j++) {
        int res = 0;
        for (auto x : a) {
          int cur = min(i, x / 3);
          int rem = x - cur - cur - cur;
          int cnt2 = min(j, (rem / 2));
          int rem2 = rem - cnt2 - cnt2;
          if (rem2 == 1 && j - cnt2 >= 2 && cur) continue;
          res = max(res, rem2);
        }
        ans = min(ans, i + j + res);
      }
    }
    cout << ans << "\n";
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  