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>
  int n;
  vector<int> v;
  int ans;
  void test(int one, int two) {
    int tans = 0;
    bitset<64> cover = 1;
    loop(i, 1, one) cover |= (cover << 1);
    loop(i, 1, two) cover |= (cover << 2);
    for (auto v : ::v) {
      roop(i, one + two + two, 0) if (cover[i] && i <= v && (v - i) % 3 == 0) {
        tans = max(tans, (v - i) / 3);
        goto ok;
      }
      return;
    ok:;
    }
    ans = min(ans, tans + one + two);
  }
  signed main() {
    int T;
    cin >> (T);
    loop(kase, 1, T) {
      cin >> n;
      v.resize(n);
      for (auto& vv : v) cin >> vv;
      ans = INT32_MAX;
      loop(i, 0, 2) loop(j, 0, 6) test(i, j);
      cout << ans << '\n';
    }
    return 0;
  }
  