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;
  #pragma GCC optimize("O3")
  #pragma GCC target("avx,avx2,sse,sse2,sse3,sse4,popcnt,fma")
  #pragma GCC optimize("unroll-loops")
  const long long int MAXN = 2e5 + 5;
  const long long int mod = 1000000007;
  const long long int N = 200005;
  void solve() {
    int n;
    cin >> n;
    vector<long long int> vec(n);
    for (int i = 0; i < n; i++) cin >> vec[i];
    vector<long long int> ans(5, 0ll);
    for (auto x : vec) {
      if (x % 3ll) {
        ans[0] = 1e18;
        break;
      } else {
        ans[0] = max(ans[0], x / 3ll);
      }
    }
    for (auto x : vec) {
      if (x % 3ll != 2ll) {
        ans[1] = max(ans[1], x / 3ll);
      } else {
        ans[1] = 1e18;
        break;
      }
    }
    ans[1] += 1ll;
    for (auto x : vec) {
      if (x % 3ll != 1ll) {
        ans[2] = max(ans[2], x / 3ll);
      } else {
        ans[2] = 1e18;
        break;
      }
    }
    ans[2] += 1ll;
    for (auto x : vec) {
      if (x % 3ll) {
        ans[3] = max(ans[3], x / 3ll);
      } else {
        ans[3] = max(ans[3], x / 3ll - 1ll);
      }
    }
    ans[3] += 2ll;
    for (auto x : vec) {
      if (x == 1ll) {
        ans[4] = 1e18;
        break;
      }
      if (x % 3ll == 1ll)
        ans[4] = max(ans[4], (x - 4ll) / 3ll);
      else {
        ans[4] = max(ans[4], x / 3ll);
      }
    }
    ans[4] += 2ll;
    sort(ans.begin(), ans.end());
    cout << ans[0] << "\n";
    return;
  }
  int main() {
    ios_base ::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  