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 N = 5e5 + 10;
  const int M = 9699690;
  const int mod = 998244353, Gi = 332748118;
  const int inf = 0x3f3f3f3f;
  const long long INF = 0x3f3f3f3f3f3f3f3fll;
  const int SIZE = 1 << 21;
  char ibuf[SIZE], *iS, *iT;
  using std::pair;
  template <class T>
  void read(T &x) {
    x = 0;
    int f = 0;
    char c = getchar();
    while (!isdigit(c)) f |= c == '-', c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
    if (f) x = -x;
  }
  int random(int i) { return rand() % i + 1; }
  template <class T>
  void ckmax(T &x, T y) {
    x = x > y ? x : y;
  }
  template <class T>
  void ckmin(T &x, T y) {
    x = x < y ? x : y;
  }
  template <class T>
  T Abs(T x) {
    return x > 0 ? x : -x;
  }
  inline int plus(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
  int qp(int d, int k) {
    int f = 1;
    while (k) {
      if (k & 1) f = (1ll * (f) * (d) % mod);
      d = (1ll * (d) * (d) % mod), k >>= 1;
    }
    return f;
  }
  int GCD(int a, int b) { return b ? GCD(b, a % b) : a; }
  int n, a[200];
  void work() {
    read(n);
    int f0 = 0, f1 = 0, f2 = 0, mx = 0;
    for (int i = (1); i <= (n); i++) {
      read(a[i]);
      ckmax(mx, a[i]);
      f0 |= a[i] % 3 == 0;
      f1 |= a[i] % 3 == 1;
      f2 |= a[i] % 3 == 2;
    }
    if (mx % 3 == 2) {
      if (f1)
        printf("%d\n", mx / 3 + 2);
      else
        printf("%d\n", mx / 3 + 1);
    }
    if (mx % 3 == 1) {
      int f3 = 0;
      for (int i = (1); i <= (n); i++) f3 |= a[i] == mx - 1, f3 |= a[i] == 1;
      if (f2 && f3)
        printf("%d\n", mx / 3 + 2);
      else
        printf("%d\n", mx / 3 + 1);
    }
    if (mx % 3 == 0) {
      if (f1 || f2)
        printf("%d\n", mx / 3 + 1);
      else
        printf("%d\n", mx / 3);
    }
  }
  int main() {
    int T = 1;
    read(T);
    while (T--) work();
    return 0;
  }
  