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;
  int getCases() {
    int cases = -1;
    enum InputType : int {
      SingleInput = 1,
      MultipleInput = 2,
      NotSet = 0,
    };
    InputType inputType = MultipleInput;
    if (inputType == SingleInput) {
      cases = 1;
    } else if (inputType == MultipleInput) {
      scanf("%d", &cases);
    } else {
      assert(false && "fix the function");
    }
    return cases;
  }
  int n;
  int ar[105];
  int big;
  void Input() {
    big = 0;
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) {
      scanf("%d", &ar[i]);
      big = max(big, ar[i]);
    }
  }
  int main() {
    int cases = getCases();
    while (cases--) {
      Input();
      int ans = INT_MAX;
      for (int a = 0; a <= 3; ++a) {
        for (int b = 0; b <= 3; ++b) {
          int c = (big - a - 2 * b + 2) / 3;
          int good = true;
          for (int i = 0; i < n; ++i) {
            int r = ceil((double)ar[i] / 3.0);
            if (ar[i] % 3 == 0) {
              if (c >= r) {
              } else if (c >= r - 1 && ((a >= 1 && b >= 1) || a >= 3)) {
              } else {
                good = false;
              }
            } else if (ar[i] % 3 == 1) {
              if (c >= r - 1 && a >= 1) {
              } else if (ar[i] > 3 && c >= r - 2 && b >= 2) {
              } else {
                good = false;
              }
            } else if (ar[i] % 3 == 2) {
              if (c >= r - 1 && (a >= 2 || b >= 1)) {
              } else {
                good = false;
              }
            }
          }
          if (good) {
            ans = min(ans, a + b + c);
          }
        }
      }
      printf("%d\n", ans);
    }
  }
  