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 main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      vector<int> arr(n);
      for (int &e : arr) cin >> e;
      int ans = *max_element(arr.begin(), arr.end());
      int ans2 = INT_MAX;
      int cond = 1;
      for (int i : arr) {
        if (i & 1) {
          cond = 0;
          break;
        }
      }
      if (cond) ans2 = ans / 2;
      int ans3 = INT_MAX;
      cond = 1;
      for (int i : arr) {
        if (i % 3 != 0) {
          cond = 0;
          break;
        }
      }
      if (cond) ans3 = ans / 3;
      int c1 = 0, c2 = 0, c3 = 0;
      int ans4 = INT_MIN;
      for (int i : arr) {
        c2 = max(c2, i / 2);
        c1 = max(c1, i - 2 * (i / 2));
      }
      ans4 = c1 + c2;
      c1 = 0, c2 = 0, c3 = 0;
      int ans5 = INT_MIN;
      for (int i : arr) {
        c3 = max(c3, i / 3);
        c1 = max(c1, i - 3 * (i / 3));
      }
      ans5 = c1 + c3;
      c1 = 0, c2 = 0, c3 = 0;
      int ans6 = INT_MIN;
      for (int i : arr) {
        if (i == 1) {
          ans6 = INT_MAX;
          break;
        }
        if (i % 3 == 0) {
          c3 = max(c3, i / 3);
        } else if ((i - 2) % 3 == 0) {
          c3 = max(c3, (i - 2) / 3);
          c1 = max(c1, (i - 3 * ((i - 2) / 3)) / 2);
        } else {
          c3 = max(c3, (i - 4) / 3);
          c1 = max(c1, (i - 3 * ((i - 4) / 3)) / 2);
        }
        ans6 = max(ans6, c1 + c3);
      }
      c1 = 0, c2 = 0, c3 = 0;
      int ans7 = INT_MIN;
      for (int i : arr) {
        c3 = max(c3, i / 3);
        int r = i - 3 * (i / 3);
        if (r == 1 and !c1) {
          c1 = 1;
        } else if (!c2) {
          c2 = 1;
        }
      }
      ans7 = c1 + c2 + c3;
      cond = 1;
      if (c1 * c2) {
        int p3 = c3 - 1;
        for (int i : arr) {
          if (i <= 3 * p3 + 3) {
            continue;
          } else {
            cond = 0;
            break;
          }
        }
      }
      if (cond and c1 * c2) ans7--;
      vector<int> v = {ans, ans2, ans3, ans4, ans5, ans6, ans7};
      cout << *min_element(v.begin(), v.end()) << endl;
    }
  }
  