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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  #pragma GCC optimization("unroll-loops")
  using namespace std;
  const long double Pi = 3.141592653;
  const long long mod = 998244353;
  long long INF = 1000000000000000000;
  struct mycmp {
    bool operator()(pair<int, int> a, pair<int, int> b) const {
      return (a.first < b.first) || (a.first == b.first && a.second > b.second);
    }
  };
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      vector<long long> a(n);
      for (int i = 0; i < n; i++) cin >> a[i];
      long long ans5 = 0, cnt = 0;
      for (int i = 0; i < n; i++) {
        if (a[i] % 3)
          ++cnt;
        else
          ans5 = max(ans5, a[i] / 3);
      }
      if (!cnt)
        cout << ans5 << "\n";
      else {
        long long x1 = 0, x2 = 0;
        long long ans = 0, ans1 = 0, ans2 = 0;
        for (int i = 0; i < n; i++) {
          ans = max(ans, a[i] / 3);
          long long y = a[i] % 3;
          if (y == 2)
            x1 = max(x1, y);
          else if (y == 1)
            x1 = max(x1, y);
        }
        ans += x1;
        x1 = 0;
        long long ans4 = 0, o = 0;
        for (int i = 0; i < n; i++) {
          ans4 = max(ans4, a[i] / 3);
          if (a[i] == 1) ans4 = INF, ++o;
          long long y = a[i] % 3;
          if (y == 2)
            x1 = max(x1, 1LL);
          else if (y == 1)
            x1 = max(x1, 2LL);
        }
        if (x1 == 2) {
          for (int i = 0; i < n; i++) {
            if (a[i] % 3 == 1) {
              long long h = max(a[i] - 4, 0LL);
              ans1 = max(ans1, h / 3);
            } else
              ans1 = max(ans1, a[i] / 3);
          }
          ans1 += 2;
          ans4 = INF;
        } else
          ans4 += 1, ans1 = INF;
        x1 = 0, x2 = 0;
        for (int i = 0; i < n; i++) {
          if (a[i] % 3 == 1) ++x1;
          if (a[i] % 3 == 2) ++x2;
          if (a[i] % 3 == 0)
            ans2 = max(ans2, a[i] / 3 - 1);
          else
            ans2 = max(ans2, a[i] / 3);
        }
        ans2 += 2;
        if (o) ans1 = INF, ans4 = INF;
        cout << min(ans1, min(ans2, min(ans, ans4))) << "\n";
      }
    }
  }
  