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>
  int int_cmp(const void *a, const void *b) { return *((int *)a) - *((int *)b); }
  int unique(int a[], int size) {
    int i, k;
    if (size == 0) return 0;
    for (i = 1; i < size; i++)
      if (a[i] == a[i - 1]) break;
    for (k = i, i++; i < size; i++)
      if (a[i] != a[i - 1]) a[k++] = a[i];
    return k;
  }
  int main(int argc, char *argv[]) {
    int i, T, n, onec, twoc, q, r, ans;
    int cost[128];
    scanf("%d", &T);
    while (T-- > 0) {
      scanf("%d", &n);
      onec = twoc = 0;
      for (i = 0; i < n; i++) {
        scanf("%d", &cost[i]);
        onec |= (cost[i] % 3) == 1;
        twoc |= (cost[i] % 3) == 2;
      }
      qsort(cost, n, sizeof(cost[0]), int_cmp);
      n = unique(cost, n);
      q = cost[n - 1] / 3;
      if (onec == 0 && twoc == 0)
        ans = q;
      else if (onec == 0 || twoc == 0)
        ans = q + 1;
      else {
        r = cost[n - 1] % 3;
        if (r == 0)
          ans = q + 1;
        else if (r == 2)
          ans = q + 2;
        else if (cost[0] == 1 || cost[n - 2] == cost[n - 1] - 1)
          ans = q + 2;
        else
          ans = q + 1;
      }
      printf("%d\n", ans);
    }
    return 0;
  }
  