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;
  long long int gcd(long long int a, long long int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int t;
    cin >> t;
    while (t--) {
      long long int n, i, j, count = 0;
      cin >> n;
      long long int arr[n + 1];
      long long int mx = 0;
      bool a = 0, b = 0;
      set<long long int> s;
      set<long long int> s1;
      for (i = 1; i <= n; i++) {
        cin >> arr[i];
        mx = max(arr[i], mx);
        s1.insert(arr[i]);
        if (arr[i] % 3 == 1) {
          s.insert(arr[i]);
          a = 1;
        } else if (arr[i] % 3 == 2)
          b = 1;
      }
      if (mx % 3 == 0) {
        if (a == 1 || b == 1)
          cout << mx / 3 + 1 << '\n';
        else
          cout << mx / 3 << '\n';
      } else {
        if (mx < 3)
          cout << a + b << '\n';
        else if (mx % 3 == 1) {
          if ((s1.find(mx - 1) != s1.end() && b == 1) ||
              (b == 1 && s1.find(1) != s1.end()))
            cout << mx / 3 + 2 << '\n';
          else
            cout << mx / 3 + 1 << '\n';
        } else if (a == 1 && b == 1)
          cout << mx / 3 + 2 << '\n';
        else
          cout << mx / 3 + 1 << '\n';
      }
    }
    return 0;
  }
  