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;
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      long long a[n + 1];
      for (long long i = 1; i <= n; i++) {
        cin >> a[i];
      }
      bool two = 0;
      bool one = 0;
      bool onepr = 0;
      long long req[n + 1];
      for (long long i = 1; i <= n; i++) {
        if (a[i] % 3 == 1) {
          one = 1;
          req[i] = a[i] / 3 + 1;
        }
        if (a[i] % 3 == 2) {
          two = 1;
          req[i] = a[i] / 3 + 1;
        }
        if (a[i] % 3 == 0) {
          req[i] = a[i] / 3;
        }
        if (a[i] == 1) {
          onepr = 1;
        }
      }
      long long ans = a[1];
      long long ind = 1;
      for (long long i = 1; i <= n; i++) {
        if (a[i] >= ans) {
          ans = a[i];
          ind = i;
        }
      }
      if (!two && !one) {
        cout << ans / 3 << endl;
        continue;
      }
      if (!onepr) {
        if (!one) {
          ans = 0;
          for (long long i = 1; i <= n; i++) {
            if (a[i] % 3 == 2) {
              a[i] -= 2;
              req[i] = a[i] / 3;
            }
            if (req[i] > ans) {
              ans = req[i];
            }
          }
          ans++;
          cout << ans << endl;
          continue;
        }
        if (!two) {
          ans = 0;
          for (long long i = 1; i <= n; i++) {
            if (a[i] % 3 == 1) {
              a[i] -= 1;
              req[i] = a[i] / 3;
            }
            if (req[i] > ans) {
              ans = req[i];
            }
          }
          ans++;
          cout << ans << endl;
          continue;
        }
        ans = 0;
        if (a[ind] % 3 == 1) {
          for (long long i = 1; i <= n; i++) {
            if (a[i] % 3 == 2) {
              a[i] -= 2;
            }
            if (a[i] % 3 == 1) {
              a[i] -= 4;
            }
            req[i] = a[i] / 3;
            if (req[i] > ans) {
              ans = req[i];
            }
          }
          ans += 2;
          cout << ans << endl;
          continue;
        }
        if (a[ind] % 3 == 0 || a[ind] % 3 == 2) {
          ans = 0;
          for (long long i = 1; i <= n; i++) {
            if (a[i] % 3 == 0) {
              a[i] -= 3;
            }
            if (a[i] % 3 == 2) {
              a[i] -= 2;
            }
            if (a[i] % 3 == 1) {
              a[i] -= 1;
            }
            req[i] = a[i] / 3;
            if (req[i] > ans) {
              ans = req[i];
            }
          }
          ans += 2;
          cout << ans << endl;
          continue;
        }
      }
      if (onepr) {
        if (!two) {
          ans = 0;
          for (long long i = 1; i <= n; i++) {
            if (a[i] % 3 == 1) {
              a[i] -= 1;
              req[i] = a[i] / 3;
            }
            if (req[i] > ans) {
              ans = req[i];
            }
          }
          ans++;
          cout << ans << endl;
          continue;
        }
        ans = 0;
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 0) {
            a[i] -= 3;
          }
          if (a[i] % 3 == 2) {
            a[i] -= 2;
          }
          if (a[i] % 3 == 1) {
            a[i] -= 1;
          }
          req[i] = a[i] / 3;
          if (req[i] > ans) {
            ans = req[i];
          }
        }
        ans += 2;
        cout << ans << endl;
      }
    }
    return 0;
  }
  