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;
  int n, a[110], g, ans;
  bitset<10> vis;
  void check(int x, int y) {
    vis.reset(), vis[g = 0] = 1;
    for (int i = 1; i <= x; ++i) vis |= vis << 1;
    for (int i = 1; i <= y; ++i) vis |= vis << 2;
    for (int i = 1, z; i <= n; ++i) {
      z = a[i] % 3;
      if (a[i] >= 3 && vis[z + 3])
        g = max(g, a[i] / 3 - 1);
      else if (vis[z])
        g = max(g, a[i] / 3);
      else
        return;
    }
    ans = min(ans, g + x + y);
  }
  void Solve_Test() {
    cin >> n, ans = 1e9;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 0; i < 3; ++i)
      for (int j = 0; j < 3; ++j) check(i, j);
    cout << ans << endl;
  }
  const bool BASE = 1;
  int T = 1;
  int main() {
    if (BASE) cin >> T;
    while (T--) Solve_Test();
    return 0;
  }
  