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;
  inline long long read() {
    long long n = 0, f = 1, ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      n = n * 10 + ch - '0';
      ch = getchar();
    }
    return n * f;
  }
  long long a[200005];
  bool kd[8];
  signed main() {
    long long t, n, sth, ans;
    bool flag;
    t = read();
    for (long long greg = 1; greg <= t; greg++) {
      n = read();
      for (long long i = 1; i <= n; i++) a[i] = read();
      ans = 1000000000;
      for (long long i = 0; i <= 1; i++) {
        for (long long j = 0; j <= 2; j++) {
          for (long long k = 0; k <= 5; k++) kd[k] = false;
          kd[0] = kd[i] = true;
          if (j >= 1) kd[i + 2] = kd[2] = true;
          if (j >= 2) kd[i + 4] = kd[4] = true;
          sth = 0;
          flag = true;
          for (long long ti = 1; ti <= n; ti++) {
            for (long long tj = 5; tj >= 0; tj--) {
              if (a[ti] >= tj && kd[tj] == true && a[ti] % 3 == tj % 3) {
                sth = max(sth, (a[ti] - tj) / 3);
                break;
              }
              if (tj == 0) flag = false;
            }
            if (flag == false) break;
          }
          if (flag == false) continue;
          ans = min(ans, sth + i + j);
        }
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  