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;
  const long long sz = 1e5 + 10;
  const long long inf = 0x3f3f3f3f;
  inline long long read() {
    long long f = 1, x = 0;
    char ch = getchar();
    for (; !isdigit(ch); ch = getchar())
      if (ch == '-') f = -1;
    for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    return f * x;
  }
  void print(long long *z, long long n) {
    for (long long i = 1; i <= n; i++) printf("%d  ", z[i]);
    puts("");
  }
  string sa, sb;
  long long cnt[sz];
  long long n, m;
  long long mx;
  vector<vector<long long> > p;
  long long z[sz];
  long long check(long long one, long long two, long long val) {
    long long res = inf;
    for (long long i = 0; i <= one; i++) {
      for (long long j = 0; j <= two; j++) {
        if (val < i + j * 2) continue;
        if ((val - i - j * 2) % 3 == 0) {
          res = min(res, one + two + (val - i - j * 2) / 3);
        }
      }
    }
    return res;
  }
  long long check(long long one, long long two) {
    long long mx3 = 0;
    for (long long i = 1; i <= n; i++) {
      long long can = check(one, two, z[i]);
      if (can == inf) return inf;
      mx3 = max(mx3, can);
    }
    return mx3;
  }
  void solve() {
    n = read();
    for (long long i = 1; i <= n; i++) {
      z[i] = read();
    }
    long long mn = inf;
    for (long long one = 0; one <= 2; one++) {
      for (long long two = 0; two <= 2; two++) {
        mn = min(mn, check(one, two));
      }
    }
    cout << mn << endl;
  }
  signed main() {
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  