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;
  const long long MAXN = 1e5 + 123;
  const long long MAX = 1000000;
  const long long MOD = 1000000007;
  const long double pi = 3.1415926536;
  bool isPrime(long long n) {
    if (n <= 1) return 0;
    if (n <= 3) return 1;
    if (n % 2 == 0 || n % 3 == 0) return 0;
    for (auto i = 5; i * i <= n; i += 6)
      if (n % i == 0 || n % (i + 2) == 0) return 0;
    return 1;
  }
  long long cc(long long ans) {
    long long cc = (ans * (ans + 1)) / 2;
    return cc;
  }
  long long gcd(long long a, long long b) {
    if (!b) return a;
    return gcd(b, a % b);
  }
  void f() {
    long long n;
    cin >> n;
    long long a[n];
    for (auto i = 0; i < n; i++) cin >> a[i];
    long long c1 = 0, c2 = 0;
    sort(a, a + n);
    long long ans = a[n - 1] / 3;
    for (auto i = 0; i < n; i++) {
      if (a[i] % 3 == 1)
        ++c1;
      else if (a[i] % 3 == 2)
        ++c2;
    }
    if ((a[n - 1] % 3 == 0) && (c1 || c2)) ++ans;
    if ((a[n - 1] % 3 == 2)) {
      ans++;
      if (c1) ++ans;
    }
    if (a[n - 1] % 3 == 1) {
      ++ans;
      if (c2) {
        for (auto i = 0; i < n; i++)
          if (a[i] == 1 || (a[i] == (a[n - 1] - 1))) {
            ++ans;
            break;
          }
      }
    }
    cout << ans << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      f();
    }
  }
  