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;
  constexpr long long N = 2000050;
  constexpr long long D = 2000010;
  constexpr long double EPS = 1e-8;
  ofstream fout;
  ifstream fin;
  int flag = 1;
  long long n, m, k;
  void solve() {
    cin >> n;
    vector<long long> v(n);
    vector<long long> t;
    for (auto &x : (v)) cin >> x, t.push_back(x);
    long long b = 0;
    for (auto &x : (v)) {
      b = max(b, x / 3);
      x %= 3;
    }
    int a = 0, c = 0;
    if (count(v.begin(), v.end(), 1)) a++;
    if (count(v.begin(), v.end(), 2)) c++;
    sort(t.begin(), t.end()),
        (t).resize(unique(t.begin(), t.end()) - (t).begin());
    reverse(t.begin(), t.end());
    b += a + c;
    if (t[0] % 3 == 1 && c && t.back() != 1) {
      if (t[1] <= t[0] - 2) b--;
    }
    if (t[0] % 3 == 0 && a && c) b--;
    cout << b << "\n";
  }
  signed main(signed argc, char *argv[]) {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cout.precision(9);
    cout.fixed;
    srand(time(0));
    fout.precision(14);
    fout.fixed;
    int tt = 1;
    cin >> tt;
    while (tt--) {
      solve();
    }
    fout.close();
    fin.close();
  }
  