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 int inf = 1e9;
  const int maxu = 200005;
  void google(int t) { cout << "Case #" << t << ": "; }
  int gcd(int a, int b) {
    if (b > a) {
      return gcd(b, a);
    }
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  int fun(int a, int b) {
    if (b == 0)
      return 1;
    else if (a == 1)
      return 1;
    int temp = fun(a, b / 2) % 1000000007;
    temp = (temp * temp) % 1000000007;
    if (b % 2 == 1) temp = (temp * a) % 1000000007;
    return temp % 1000000007;
  }
  int inv(int a, int b) { return (a * fun(b, 1000000007 - 2)) % 1000000007; }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int i, j, k = 0, n, m, l = 0;
    int t = 1;
    int T = 1;
    cin >> T;
    for (t = 1; t <= T; t++) {
      cin >> n;
      int a[n];
      int maxi = 0;
      for (i = 0; i < n; i++) {
        cin >> a[i];
        maxi = max(maxi, a[i]);
      }
      int c3max = (maxi / 3) + 3;
      int ans = inf;
      for (int c1 = 0; c1 <= 4; c1++) {
        for (int c2 = 0; c2 <= 4; c2++) {
          for (int c3 = max(0, c3max - 6); c3 <= c3max; c3++) {
            int cnt = c1 + c2 + c3;
            if (cnt >= ans) continue;
            int can = 1;
            for (i = 0; i < n; i++) {
              int can2 = 0;
              for (j = 0; j <= c1; j++) {
                for (k = 0; k <= c2; k++) {
                  int x = a[i] - j - k * 2;
                  if (x == 0) {
                    can2 = 1;
                    break;
                  }
                  if (x < 0) continue;
                  if ((x % 3 == 0) && (x <= 3 * c3)) {
                    can2 = 1;
                    break;
                  }
                }
                if (can2 == 1) break;
              }
              if (can2 == 0) {
                can = 0;
                break;
              }
            }
            if (can) ans = cnt;
          }
        }
      }
      cout << ans << "\n";
    }
    return 0;
  }
  