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 int N = 110;
  int t, n;
  int a[N];
  int now[20];
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> t;
    while (t--) {
      cin >> n;
      for (int i = 1; i <= n; ++i) cin >> a[i];
      int ans = 1e9;
      for (int cnt1 = 0; cnt1 <= 6; ++cnt1)
        for (int cnt2 = 0; cnt2 <= 3; ++cnt2) {
          for (int i = 0; i <= 12; ++i) now[i] = 0;
          now[0] = 1;
          for (int i = 1; i <= cnt1; ++i)
            for (int j = 12; j >= 1; --j) now[j] |= now[j - 1];
          for (int i = 1; i <= cnt2; ++i)
            for (int j = 12; j >= 2; --j) now[j] |= now[j - 2];
          int maxs = 0;
          for (int i = 1; i <= n; ++i) {
            int flag = 0;
            for (int j = 12; j >= 0; --j)
              if (now[j])
                if (j <= a[i] && (a[i] - j) % 3 == 0) {
                  flag = 1;
                  maxs = max(maxs, (a[i] - j) / 3);
                  break;
                }
            if (flag == 0) maxs = 1e9;
          }
          ans = min(ans, maxs + cnt1 + cnt2);
        }
      cout << ans << "\n";
    }
  }
  