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;
  bool che(int k, vector<int> &a) {
    int n = a.size();
    for (int i = 0; i < 3; i++) {
      for (int j = 0; j < 3; j++) {
        int x = k - i - j;
        if (x < 0) continue;
        bool chek = 1;
        for (int p = 0; p < n; p++) {
          bool h = 0;
          for (int l = 0; l <= i; l++) {
            for (int w = 0; w <= j; w++) {
              int t = a[p] - l - 2 * w;
              if ((t - 3 * x) <= 0 and t >= 0 and (t % 3 == 0)) h = 1;
            }
          }
          if (!h) {
            chek = 0;
            break;
          }
        }
        if (chek) return 1;
      }
    }
    return 0;
  }
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    sort(a.rbegin(), a.rend());
    int k = a[0] / 3;
    if (a[0] % 3 != 0) k++;
    int l = 1, r = a[0] + 1;
    while (l != r) {
      int m = (l + r) / 2;
      if (che(m, a))
        r = m;
      else
        l = m + 1;
    }
    cout << l << "\n";
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int tt;
    cin >> tt;
    while (tt--) solve();
    return 0;
  }
  