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;
  long long M1 = pow(10, 9) + 7, M2 = 998244353, NX = 9223372036854775807,
            LN = 1e18;
  long long binpow(long long x, long long n) {
    long long res = 1;
    while (n) {
      if (n % 2) res = (res * x);
      n /= 2;
      x = (x * x);
    }
    return res;
  }
  long long modpow(long long x, long long n, long long M) {
    x = x % M;
    long long res = 1;
    while (n) {
      if (n % 2) res = (res * x) % M;
      n /= 2;
      x = (x * x) % M;
    }
    return res;
  }
  long long gcd(long long a, long long b) {
    if (!b) return a;
    return gcd(b, a % b);
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long tc = 1;
    cin >> tc;
    while (tc--) {
      long long n;
      cin >> n;
      vector<long long> v(n);
      long long i, m = 0, cnt[3] = {};
      for (i = 0; i < n; i++) {
        cin >> v[i];
        m = max(m, v[i]);
        cnt[v[i] % 3] = 1;
      }
      long long f = 0, g = 0;
      for (i = 0; i < n; i++) {
        if (v[i] == 1) f = 1;
        if (v[i] == m - 1) g = 1;
      }
      if (m % 3 == 0)
        cout << m / 3 + (cnt[1] | cnt[2]) << "\n";
      else if (m % 3 == 1) {
        if (!cnt[2] || (cnt[2] && !f && !g))
          cout << m / 3 + 1 << "\n";
        else
          cout << m / 3 + 2 << "\n";
      } else
        cout << m / 3 + 1 + cnt[1] << "\n";
    }
    return 0;
  }
  