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 double scale = 1e+6;
  const long long mod = 1000000007;
  int ispower(long long x, long long y) {
    long long res = 1;
    while (y > 0) {
      if (y & 1) res = res * x;
      y = y >> 1;
      x = x * x;
    }
    return res;
  }
  const long long N = 100001;
  using namespace std;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) cin >> a[i];
    sort((a).begin(), (a).end());
    long long m = a[n - 1] / 3, ans = INT_MAX;
    for (long long i = 0; i <= 1; i++) {
      for (long long j = 0; j <= 2; j++) {
        for (long long k = max(1ll * 0, m - 4); k <= m; k++) {
          long long t = i + j + k, c = 0;
          long long f = 0;
          for (auto it : a) {
            long long flag = 0;
            for (long long x = 0; x <= i; x++)
              for (long long y = 0; y <= j; y++) {
                long long s = it - x - 2 * y;
                if (s >= 0 && s % 3 == 0 && s / 3 <= k) {
                  c++;
                  flag = 1;
                  break;
                }
              }
            if (flag == 0) {
              f = 1;
              break;
            }
          }
          if (f == 0) ans = min(ans, t);
        }
      }
    }
    cout << ans << endl;
  }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cout << fixed << setprecision(25);
    cerr << fixed << setprecision(10);
    auto start = std::chrono::high_resolution_clock::now();
    solve();
    auto stop = std::chrono::high_resolution_clock::now();
    auto duration =
        std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);
    return 0;
  }
  