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;
  void judge() {}
  void fastio() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  long long int p1 = 1e9 + 7;
  long long int p2 = 998244353;
  struct segtree {
    vector<int> v;
    int n, n2;
    void init(int x) {
      n = x;
      int a = 1;
      while (a <= x) a *= 2;
      a *= 2;
      n2 = a;
      v.assign(n2, 0);
    }
    int return_max(int curr, int r2, int l, int r) {
      if (l == r) {
        return v[curr];
      }
      int mid = (r + l) / 2;
      if (r2 <= mid) return return_max(2 * curr + 1, r2, l, mid);
      return max(v[2 * curr + 1], return_max(2 * curr + 2, r2, mid + 1, r));
    }
    int return_max(int r) {
      if (r < 0) return 0;
      return return_max(0, r, 0, n - 1);
    }
    void update(int curr, int pos, int val, int l, int r) {
      if (l == r) {
        v[curr] = max(v[curr], val);
        return;
      }
      int mid = (l + r) / 2;
      if (pos <= mid)
        update(2 * curr + 1, pos, val, l, mid);
      else
        update(2 * curr + 2, pos, val, mid + 1, r);
      v[curr] = max(v[2 * curr + 1], v[2 * curr + 2]);
    }
    void update(int pos, int val) { update(0, pos, val, 0, n - 1); }
    int ans() {
      int a = 0;
      for (auto val : v) a = max(a, val);
      return a;
    }
  };
  bool compare(pair<long long int, long long int>& a,
               pair<long long int, long long int>& b) {
    if (a.first != b.first) return a.first > b.first;
    return a.second < b.second;
  }
  int find_set(vector<int>& parent, int x) {
    if (parent[x] == x) return x;
    return parent[x] = find_set(parent, parent[x]);
  }
  void union_set(vector<int>& parent, int x, int y) {
    x = find_set(parent, x);
    y = find_set(parent, y);
    if (x != y) {
      parent[x] = y;
    }
  }
  vector<long long int> primes;
  void pre() {
    vector<long long int> v(100000);
    for (long long int i = 2; i <= 100000; i++) {
      if (v[i] == 0) {
        primes.push_back(i);
        for (long long int j = i * i; j <= 10000; j += i) {
          v[j] = 1;
        }
      }
    }
  }
  long long int query(long long int a, long long int b, long long int c) {
    cout << "? " << a << " " << b << " " << c << "\n";
    cout.flush();
    long long int l;
    cin >> l;
    return l;
  }
  void solve2() {
    long long int n;
    cin >> n;
    vector<long long int> v;
    long long int a = 0;
    long long int b = 0;
    long long int c = 0;
    long long int d = 0;
    for (int i = 0; i < n; i++) {
      long long int l;
      cin >> l;
      v.push_back(l);
      a += ((l % 3) != 0);
      b += (l == 1);
      c += ((l % 3) == 2);
      d += ((l % 3) == 1);
    }
    sort(v.begin(), v.end());
    set<long long int> ss;
    for (int i = 0; i < n; i++) {
      if (v[i] >= ((v[n - 1] / 3) * 3) && v[i] != v[n - 1]) ss.insert(v[i]);
    }
    if (v[n - 1] % 3 == 0) {
      if (a != 0) {
        cout << (v[n - 1] / 3) + 1;
      } else
        cout << (v[n - 1] / 3);
      return;
    } else if (v[n - 1] % 3 == 1) {
      if (b != 0) {
        if (c != 0) {
          cout << (v[n - 1] / 3) + 2;
        } else
          cout << (v[n - 1] / 3) + 1;
      } else {
        auto it = ss.find(v[n - 1] - 1);
        long long int e = 0;
        if (it != ss.end()) {
          e = 1;
        }
        if (c != 0 && e == 1) {
          cout << (v[n - 1] / 3) + 2;
        } else
          cout << (v[n - 1] / 3) + 1;
      }
    } else {
      if (d != 0) {
        cout << (v[n - 1] / 3) + 2;
      } else
        cout << (v[n - 1] / 3) + 1;
    }
  }
  int main() {
    judge();
    fastio();
    int t = 1;
    cin >> t;
    for (int i = 0; i < t; i++) {
      solve2();
      cout << " \n";
    }
    cerr << "Time : " << ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
    return 0;
  }
  