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;
  using ll = long long;
  using ii = pair<int, int>;
  const int MAXN = 1e5;
  int solve(int ones, int twos, vector<int>& v, int n) {
    int ans = 0;
    for (int i = 0; i < n; i++) {
      if (v[i] % 3 == 0) {
        int cand = v[i] / 3;
        if (ones and twos) {
          cand--;
        }
        ans = max(ans, cand);
      } else if (v[i] % 3 == 1) {
        if (ones) {
          ans = max(ans, v[i] / 3);
        } else if (v[i] == 1) {
          return 1e9;
        } else if (twos > 1) {
          ans = max(ans, (v[i] - 4) / 3);
        } else {
          return 1e9;
        }
      } else {
        if (ones > 1) {
          ans = max(ans, v[i] / 3);
        } else if (twos) {
          ans = max(ans, v[i] / 3);
        } else {
          return 1e9;
        }
      }
    }
    return ones + twos + ans;
  }
  int tc() {
    int n;
    scanf("%d", &n);
    vector<int> v(n);
    for (int i = 0; i < n; i++) {
      scanf("%d", &v[i]);
    }
    sort(v.begin(), v.end());
    int ans = 1e9;
    for (int i = 0; i < 3; i++) {
      for (int j = 0; j < 3; j++) {
        int cand = solve(i, j, v, n);
        ans = min(ans, cand);
      }
    }
    return ans;
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      printf("%d\n", tc());
    }
  }
  