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>
  #pragma GCC optimize("O3", "Ofast", "no-stack-protector", "unroll-loops", \
                       "omit-frame-pointer", "inline")
  #pragma GCC option("arch=native", "tune=native", "no-zero-upper")
  using namespace std;
  struct st {
    long long A, B, C;
  };
  bool ok(st arr, long long N) {
    for (long long i = 0; i <= arr.A; i++) {
      for (long long j = 0; j <= arr.B; j++) {
        if (i * 1 + j * 2 <= N && (N - (i * 1 + j * 2)) % 3 == 0 &&
            (N - (i * 1 + j * 2)) / 3 <= arr.C) {
          return true;
        }
      }
    }
    return false;
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    vector<st> arr;
    for (long long i = 0; i <= 6; i++) {
      for (long long j = 0; j <= 6; j++) {
        for (long long k = 0; k <= 6; k++) {
          if (i + j + k <= 6) arr.push_back({i, j, k});
        }
      }
    }
    long long T;
    cin >> T;
    while (T--) {
      long long N;
      cin >> N;
      vector<long long> vect;
      long long tot = 0;
      for (long long i = 0; i < N; i++) {
        long long n;
        cin >> n;
        vect.push_back(n);
        tot = max(tot, n / 3);
      }
      tot = max(0ll, tot - 4);
      long long res = 1e18;
      for (auto x : arr) {
        x.C += tot;
        long long cont = 0;
        for (long long i = 0; i < N; i++) {
          if (ok(x, vect[i])) cont++;
        }
        if (cont == N) {
          res = min(res, x.A + x.B + x.C);
        }
      }
      cout << res << '\n';
    }
    return 0;
  }
  