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 t;
  long long n;
  long long a[110];
  int main(void) {
    scanf("%lld", &t);
    while (t--) {
      scanf("%lld", &n);
      for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
      long long chk[10] = {
          0,
      };
      for (long long i = 1; i <= n; i++) chk[a[i] % 3] = 1;
      long long ans = 0;
      if (!chk[1] && !chk[2]) {
        for (long long i = 1; i <= n; i++) ans = max(a[i] / 3, ans);
      } else if (chk[1] && !chk[2]) {
        for (long long i = 1; i <= n; i++) ans = max(ans, a[i] / 3);
        ans++;
      } else if (!chk[1] && chk[2]) {
        for (long long i = 1; i <= n; i++) ans = max(ans, a[i] / 3);
        ans++;
      } else {
        long long ff = 0;
        for (long long i = 1; i <= n; i++) {
          if (a[i] == 1) ff = 1;
        }
        if (ff) {
          for (long long i = 1; i <= n; i++) {
            if (a[i] % 3 == 0)
              ans = max(ans, a[i] / 3 - 1);
            else
              ans = max(ans, a[i] / 3);
          }
          ans += 2;
        } else {
          long long hubo2 = 0;
          for (long long i = 1; i <= n; i++) {
            if (a[i] % 3 == 0)
              hubo2 = max(hubo2, a[i] / 3 - 1);
            else
              hubo2 = max(hubo2, a[i] / 3);
          }
          hubo2 += 2;
          long long hubo = 0;
          for (long long i = 1; i <= n; i++) {
            if (a[i] % 3 == 1)
              hubo = max(hubo, (a[i] - 4) / 3);
            else
              hubo = max(hubo, a[i] / 3);
          }
          hubo += 2;
          for (long long i = 1; i <= n; i++) {
            if (a[i] % 3 == 1)
              ans = max(ans, (a[i] - 4) / 3);
            else if (a[i] % 3 == 0)
              ans = max(ans, a[i] / 3 - 2);
            else
              ans = max(ans, (a[i] - 2) / 3);
          }
          ans += 3;
          ans = min(ans, hubo);
          ans = min(ans, hubo2);
        }
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  