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("O3,no-stack-protector,unroll-loops")
  #pragma GCC target("avx2,fma")
  using namespace std;
  const long long inf = 2e9, N = 1e6 + 10, mod = 1e9 + 7, pr = 998244353;
  int a[N];
  int cnt1, cnt2, cnt3;
  int ans1, ans2, ans3;
  int ans_t;
  bool g;
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ;
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      ans_t = inf;
      cnt1 = 0, cnt2 = 0, cnt3 = 0;
      ans1 = 0, ans2 = 0, ans3 = 0;
      g = 0;
      for (int i = 0; i < n; i++) {
        cin >> a[i];
        if (a[i] == 1) g = 1;
        if (a[i] % 3 == 0)
          cnt3 = max(cnt3, a[i]);
        else if (a[i] % 3 == 2)
          cnt2 = max(cnt2, a[i]);
        else
          cnt1 = max(cnt1, a[i]);
      }
      if (g == 0) {
        if (cnt2 > 0) {
          ans2 = 1;
          ans3 = cnt2 / 3;
        }
        if (cnt1 > 0) {
          ans2 = 2;
          ans3 = max(ans3, (cnt1 - 4) / 3);
        }
        if (cnt3 > 0) {
          if (ans1 == 0 || ans2 == 0)
            ans3 = max(ans3, (cnt3 / 3));
          else
            ans3 = max(ans3, (cnt3 - 3) / 3);
        }
        ans_t = ans1 + ans2 + ans3;
      }
      ans1 = 0, ans2 = 0, ans3 = 0;
      if (cnt2 > 0) {
        ans2 = 1;
        ans3 = cnt2 / 3;
      }
      if (cnt1 > 0) {
        ans1 = 1;
        ans3 = max(ans3, (cnt1 - 1) / 3);
      }
      if (cnt3 > 0) {
        if (ans1 == 0 || ans2 == 0)
          ans3 = max(ans3, (cnt3 / 3));
        else
          ans3 = max(ans3, (cnt3 - 3) / 3);
      }
      ans_t = min(ans_t, ans1 + ans2 + ans3);
      ans1 = 0, ans2 = 0, ans3 = 0;
      if (cnt2 > 0) {
        ans1 = 2;
        ans3 = cnt2 / 3;
      }
      if (cnt1 > 0) {
        if (ans1 == 0) ans1 = 1;
        ans3 = max(ans3, (cnt1 - 1) / 3);
      }
      if (cnt3 > 0) {
        ans3 = max(ans3, (cnt3 / 3));
      }
      ans_t = min(ans_t, ans1 + ans2 + ans3);
      cout << ans_t << "\n";
    }
    return 0;
  }
  