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;
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <typename T>
  void _do(T x) {
    cerr << x << "\n";
  }
  template <typename T, typename... U>
  void _do(T x, U... y) {
    cerr << x << ", ";
    _do(y...);
  }
  const int MOD1 = 1e9 + 7;
  const int MOD2 = 998244353;
  const long long INF = 3e18;
  const long double PI = 3.14159265358979323846;
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long fpow(long long a, long long b, long long m) {
    if (!b) return 1;
    long long ans = fpow(a * a % m, b / 2, m);
    return (b % 2 ? ans * a % m : ans);
  }
  long long inv(long long a, long long m) { return fpow(a, m - 2, m); }
  const int N = 5e5 + 1;
  long long a[105];
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      for (int i = 1; i <= n; i++) cin >> a[i];
      long long ans = INF;
      for (int x = 0; x < 4; x++) {
        for (int y = 0; y < 4; y++) {
          bool can = 1;
          long long mx = 0;
          for (int i = 1; i <= n; i++) {
            long long tmp = INF;
            for (int j = 0; j < x + 1; j++) {
              for (int k = 0; k < y + 1; k++) {
                long long b = a[i] - j - k * 2;
                if (b < 0 || b % 3 != 0) continue;
                tmp = min(tmp, b / 3);
              }
            }
            if (tmp == INF) {
              can = 0;
              break;
            }
            mx = max(mx, tmp);
          }
          if (can) ans = min(ans, mx + x + y);
        }
      }
      cout << ans << "\n";
    }
  }
  