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 fast_mod_pow(int x, int y, int mod) {
    int res = 1;
    while (y > 0) {
      if ((y & 1) == 1) {
        res = ((long long)res * x) % mod;
      }
      y >>= 1;
      x = ((long long)x * x) % mod;
    }
    return res;
  }
  int inverse_mod(int n, int p) { return fast_mod_pow(n, p - 2, p); }
  int divide_mod(int a, int b, int mod) {
    return ((long long)a * inverse_mod(b, mod)) % mod;
  }
  void solve() {
    int n;
    cin >> n;
    vector<int> A(n);
    int maxi = -1;
    for (int i = 0; i < n; ++i) {
      cin >> A[i];
      maxi = max(maxi, A[i]);
    }
    int best = 2e9;
    maxi /= 3;
    for (int ones = 0; ones <= 3; ++ones) {
      for (int twos = 0; twos <= 3; ++twos) {
        for (int threes = max(0, maxi - 3); threes <= maxi + 2; ++threes) {
          int k = ones + twos + threes;
          if (k >= best) {
            break;
          }
          bool valid = true;
          for (int i = 0; i < n; ++i) {
            int a = A[i];
            bool found = false;
            for (int x = 0; x <= ones; ++x) {
              for (int y = 0; y <= twos; ++y) {
                int rem = a - x - 2 * y;
                if (rem >= 0 and rem % 3 == 0 and rem / 3 <= threes) {
                  found = true;
                  break;
                }
              }
            }
            if (not found) {
              valid = false;
              break;
            }
          }
          if (valid) {
            best = k;
          }
        }
      }
    }
    cout << best << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  