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>
  const int maxn = 105;
  int n;
  int a[maxn];
  void solve() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", a + i);
    }
    int ans = 1000000000;
    for (int i = 0; i <= 10; i++) {
      for (int j = 0; j <= 10; j++) {
        int max3 = 0;
        for (int k = 1; k <= n; k++) {
          int temp = 1000000000;
          for (int p1 = 0; p1 <= 2; p1++) {
            for (int p2 = 0; p2 <= 2; p2++) {
              if (p1 > i || p2 > j) continue;
              int rem = a[k] - p1 - p2 - p2;
              if (rem < 0 || rem % 3 != 0) continue;
              temp = std::min(temp, rem / 3);
            }
          }
          max3 = std::max(max3, temp);
        }
        ans = std::min(ans, max3 + i + j);
      }
    }
    printf("%d\n", ans);
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
      solve();
    }
    return 0;
  }
  